Question 13 Marks
A compound is formed by $24g$ of $X$ and $64g$ of oxygen. If the atomic mass of $X=12$ and $O=16$, calculate the simplest formula of the compound.
AnswerMass of $X$ in the given compound $=24 g$
Mass of oxygen in the given compound $=64 g$
So total mass of the compound $=24+64=88 g$
$\%$ of $X$ in the compound $=24 / 88100=27.3 \%$
$\%$ of oxygen in the compound=64/88 $100=72.7 \%$
Element \% At. Mass Atomic ratio Simplest ratio
$X 27.3 : 12 = 27.3/12 = 2.27 : 1$
$O 72.7 : 16 = 72.2/16 = 4.54 : 2$
So simplest formula = $XO_2$
View full question & answer→Question 23 Marks
Urea $[CO(NH_2)_2]$ is an important nitrogenous fertilizer. Urea is sold in 50 kg sacks. What mass of nitrogen is in one sack of urea?
AnswerMolecular mass of urea $= 12 + 16+2(14+2) = 60g$
$60g$ of urea contains nitrogen $= 28g$
So, in 50g of urea, nitrogen present $= 23.33 g$
50 kg of urea contains nitrogen $= 23.33kg$
View full question & answer→Question 33 Marks
Some of the fertilizers are sodium nitrate $NaNO _3$, ammonium sulphate $\left( NH _4\right)_2 SO _4$ and urea $CO \left( NH _2\right)_2$. Which of these contains the highest percentage of nitrogen?
Answer$\%$ of $N$ in $NaNO_3= `14/85 xx 100 = 16.47\%`$
$\%$ of $N$ in $(NH_4)_2SO_4 = `14/132 xx 100 = 21.21\%$`
$\%$ of $N$ in $CO(NH_2)_2 = `14/60 xx 100 = 46.66\%`$
So, highest percentage of $N$ is in urea.
View full question & answer→Question 43 Marks
(a) Calculate the number of moles and the number of molecules present in $1.4 \ g$ of ethylene gas. What is the volume occupied by the same amount of ethylene?
(b) What is the vapour density of ethylene?
Answer(a) The molecular mass of ethylene $\left( C _2 H _4\right)$ is 28 g
No. of moles $=1.4 / 28=0.05$ moles
No. of molecules $=6.023 \times 10^{23} \times 0.05=3 \times 10^{22}$ molecules
Volume $=22.4 \times 0.05=1.12$ litres
(b) Molecular mass $=2 XV.D$
S0, $V.D = 28/2 = 14$
View full question & answer→Question 53 Marks
From the equation for burning of hydrogen and oxygen $2 H _2+ O _2 \longrightarrow 2 H _2 O$ (Steam)
Write down the number of mole (or moles) of steam obtained from 0.5 moles of oxygen.
Answer$
3 CuO +2 NH _3 \longrightarrow 3 Cu +3 H _2 O + N _2
$
The molecular mass of $3 Cu 3 CuO =(240) g$
Molecular mass of $2 NH _3=2 \times 22.4=44.8 dm ^3$
Molecular mass of $3 Cu =(192) g$
$240 g$ of $CuO$ requires $44.8 dm ^3$ of $NH _3$
$\therefore 120 g$ of CuO will require $\frac{120 \times 44.8}{240}$
$=22.4 dm ^3$
View full question & answer→Question 63 Marks
A sample of ammonium nitrate when heated yields 8.96 litres of steam (measure at $S.T.P$.)
$NH_4NO_3 \rightarrow N_2O + 2H_2O$
(i) What volume of dinitrogen oxide is produced at the same time as 8.96 litres of steam?
(ii) What mass of ammonium nitrate should be heated to produce 8.96 litres of steam? (Relative molecular mass of ammonium nitrate is $80$)
(iii) Determine the percentage of oxygen in ammonium nitrate $(O = 16).$
Answer(a) $NH _4 NO _3 \rightarrow N_2 O +2 H _2 O$
1mole1mole2mole
1 V1 V2 V
44.8 litres of water produced by $=22.4$ litres of $NH _4 NO _3$
So, 8.96 litres will be produced by $=22.4 \times 8.96 / 44.8$ $=4.48$ litres of $NH _4 NO _3$
So, 4.48 litres of $N _2 O$ is produced.
(i) 44.8 litre $H _2 O$ is produced by $=80 g$ of $NH _4 NO _3$
So, 8.96 litre $H _2 O$ will be produced by $=80 \times 8.96 / 44.8$ $=16 g NH _4 NO _3$
(iii) $\%$ of O in $NH _4 NO _3=3 \times 16 / 80=60 \%$
View full question & answer→Question 73 Marks
The equation given below relates the manufacture of sodium carbonate (molecular weight of $\left. Na _2 CO _3=106\right)$
- $\ce{NaCl + NH3 + CO2 + H2O-> NaHCO3 + NH4Cl}$
- $\ce{2NaHCO3-> Na2CO3 + H2O + CO2}$
Equations (1) and (2) are based on the production of 21.2 g of sodium carbonate.
(a) What mass of sodium hydrogen carbonate must be heated to give 21.2 g of sodium carbonate?
(b) To produce the mass of sodium hydrogen carbonate calculate in (a), what volume of carbon dioxide, measured at S.T.P. would be required? Answer(a) $\ce{NaCl + NH3 + CO2 + H2O-> NaHCO3 + NH4Cl}$
$\ce{2NaHCO3-> Na2CO3 + H2O + CO2}$
From equation:
106 g of $Na _2 CO _3$ is produced by $=168 g$ of $NaHCO _3$
So, 21.2 g of $Na _2 CO _3$ will be produced by $=168 \times 21.2 / 106$ $=33.6 g$ of $NaHCO _3$
(b) For 84 g of $NaHCO _3$, required volume of $CO _2=22.4$ litre
So, for 33.6 g of $NaHCO _3$, required volume of $CO _2=22.4 \times 33.6 / 84$ $=8.96$ litre
View full question & answer→Question 83 Marks
(a) A flask contains 3.2 g of sulphur dioxide. Calculate the following:
(i) The moles of sulphur dioxide present in the flask.
(ii) The number of molecules of sulphur dioxide present in the flask.
(iii) The volume occupied by $3.2$ g of sulphur dioxide at $S.T.P.$
$(S= 32, O= 16)$
Answer(a)
(i) The no. of moles of $SO _2=3.2 / 64=0.05$ moles
(ii) In 1 mole of $SO _2$, no. of molecules present $=6.02 \times 10^{23}$
So, in 0.05 moles, no. of molecules $=6.02 \times 10^{23} \times 0.05$
$=3.0 \times 10^{22}$
(iii) The volume occupied by 64 g of $SO _2=22.4 dm ^3$
3.2 g of $SO _2$ will be occupied by volume $=22.4 \times 3.2 / 64=1.12 dm ^3$
View full question & answer→Question 93 Marks
$10 g$ of a mixture of sodium chloride and anhydrous sodium sulphate is dissolved in water. An excess of barium chloride solution is added and 6.99 g of barium sulphate is precipitated according to the equation given below:
$Na_2SO_4 + BaCl_2 \rightarrow BaSO_4 + 2NaCl$
Calculate the percentage of sodium sulphate in the original mixture.
Answer$Na_2 SO_4+BaCl_2 \rightarrow BaSO_4+2 NaCl$
Molecular mass of $BaSO _4=233 g$
Now, 233 g of $BaSO _4$ is produced by $Na _2 SO _4=142 g$
So, $6.99 g BaSO _4$ will be produced by $=6.99 \times 142 / 233=4.26$
The percentage of $Na _2 SO _4$ in original mixture $=4.26 \times 100 / 10$ $=42.6 \%$
View full question & answer→Question 103 Marks
What volume of oxygen is required to burn completely a mixture of $22.4 dm ^3$ of methane and $11.2 dm ^3$ of hydrogen into carbon dioxide and steam?
$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$
$2H_2 + O_2 \rightarrow 2H_2O$
Answer$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$
$1 V 2 V 1 V 2 V$
From equation:
$22.4$ litres of methane requires oxygen = 44.8 litres $O_2$
$2H_2 + O_2 \rightarrow 2H_2O$
$2 V 1 V 2 V$
From equation,
$44.8$ litres hydrogen requires oxygen $= 22.4$ litres $O_2$
So, $11.2$ litres will require $= 22.4 x 11.2/44.8 = 5.6$ litres
Total volume $= 44.8 + 5.6 = 50.4$ litres
View full question & answer→Question 113 Marks
Samples of the gases $O _2, N_2, CO _2$ and CO under the same conditions of temperature and pressure contain the same number of molecules x . The molecules of oxygen occupy V litres and have a mass of 8 g under the same conditions of temperature and pressure
What is the volume occupied by:
(a) x molecules of $N _2$
(b) $3 x$ molecules of CO
(c) What is the mass of $CO _2$ in grams?
(d) In answering the above questions, which law have you used?
Answer(a) Volume of $O _2= V$
Since $O _2$ and $N _2$ have same no. of molecules $=x$ so, the volume of $N _2= V$
(b) 3 x molecules means 3 V volume of CO
(c) 32 g oxygen is contained in $=44 g$ of $CO _2$
So, 8 g oxygen is contained in $=44 \times 8 / 32=11 g$
(d) Avogadro's law is used in the above questions.
View full question & answer→Question 123 Marks
The reaction: $4N_2O + CH_4 \rightarrow CO_2 + 2H_2O + 4N_2$ takes place in the gaseous state. If all volumes are measured at the same temperature and pressure, calculate the volume of dinitrogen oxide ($N_2O$) required to give $150 cm^3$ of steam.
Answer$4N_2O + CH_4 \rightarrow CO_2 + 2H_2O + 4N_2$
$4 V 1 V 1 V 2 V 4 V$
2 x 22400 litre steam is produced by $N_2O = 4 x 22400 cm^3$
So, $150 cm^3$ steam will be produced by $= 4 \times 22400 \times 150/2 x 22400$
$= 300 cm^3N_2O$
View full question & answer→Question 133 Marks
From the equation
$3Cu + 8HNO_3 \rightarrow 3Cu (NO_3)_2+ 4H_2O + 2NO$
(Atomic mass $Cu = 64, H = 1, N = 14,O = 16)$
Calculate:
(a) Mass of copper needed to react with $63g$ of $HNO_3$
(b) The volume of nitric oxide at S.T.P. that can be collected.
Answer$3Cu + 8HNO_3 \rightarrow 3Cu (NO_3)_2+ 4H_2O + 2NO$
$1V 8V 3V 2V$
Mol. Mass of $8 HNO _3=8 \times 63=504 g$
(a) For $504 g HNO _3, Cu$ required is $=192 g$
So, for $63 g HNO _3 Cu$ required $=192 \times 63 / 504=24 g$
(b) 504 g of $HNO _3$ gives $=2 \times 22.4$ litre volume of NO
So, 63 g of $HNO _3$ gives $=2 \times 22.4 \times 63 / 504=5.6$ litre of NO
View full question & answer→Question 143 Marks
$112 cm^3$ of a gaseous fluoride of phosphorus has a mass of $0.63 g$. Calculate the relative molecular mass of the fluoride. If the molecule of the fluoride contains only one atom of phosphorus, then determine the formula of the phosphorus fluoride.$ [ F=19, P=31].$
Answer$112 cm^3$ of gaseous fluoride has mass $=0.63 g$
so, $22400 cm^3$
will have mass $=0.63 \times 22400 / 112$ $=126 g$
The molecular mass $=$ At mass $P+$
At. mass of $F$ $126=31+$ At. Mass of $F$
So, At. Mass of $F =95 g$
But, at. mass of $F =19$ so $95 / 19=5$
Hence, there are 5 atoms of F so the molecular formula $= PF _5$
View full question & answer→Question 153 Marks
Concentrated nitric acid oxidizes phosphorus to phosphoric acid according to the following equation:
$P+ 5HNO_3 \rightarrow H_3PO_4+ 5NO_2 + H_2O$
If 6.2g of phosphorus was used in the reaction calculate:
(a) The number of moles of phosphorus taken and mass of phosphoric acid
formed.
(b) mass of nitric acid will be consumed at the same time?
(c) The volume of steam produced at the same time if measured at 760 mm Hg pressure and $273^0C?$
Answer(a) 1 mole of phosphorus atom $= 31$ g of phosphorus
$31 g$ of $P = 1$ mole of $P$
$6.2g$ of $P = `(6.2 xx 1)/31` = 0.2$ mole of $P$
(b) $31 g$ $P$ reacts with $HNO_3= 315 g$
so, $6.2 g$ P will react with $HNO_3 = 315 \times 6.2/31 = 63 g$
(c)
Moles of steam formed from $31g$ phosphorus $= 18g/18g = 1mol$
Moles of steam formed from $6.2$ g phosphorus $= 1mol/31g 6.2 = 0.2$ mol
Volume of steam produced at $STP = 0.2 \times 22.4 l/MOL= 4.48$ litre
Since the pressure (760mm) remains constant , but the temperature
$(273+273)=546$ is double, the volume of the steam also gets doubled
So,Volume of steam produced at 760mm Hg and $273^0C = 4.48 \times 2 = 8.96$litre
View full question & answer→Question 163 Marks
Ammonia burns in oxygen and the combustion, in the presence of a catalyst. May be represented by
$2 NH _3+2 \frac{1}{2} O _2 \longrightarrow 2 NO +3 H _2 O $
${[ H =1, N =14, O =16]}$
What mass of steam is produced when $1.5 g$ of nitrogen monoxide is formed?
AnswerFrom equation, $2 NH _3+2 \frac{1}{2} O _2 \longrightarrow 2 NO +3 H _2 O$
When $60 g$ NO is formed, the mass of steam produced $=54 g$
So, $1.5 g NO$ is formed, the mass of steam produced $=54 \times 1.5 / 60$
$=1.35\ g$
View full question & answer→Question 173 Marks
Hydrogen sulphide gas burns in oxygen to yield $12.8 g$ of sulphur dioxide gas as under :
$2H_2S + 3O_2 \rightarrow 2H_2O + 2SO_2$
Calculate the volume of hydrogen sulphide at S.T.P. Also, calculate the volume of oxygen required at $S.T.P.$ which will complete the combustion of hydrogen sulphide determined in (litres).
Answer$2H_2S + 3O_2 \rightarrow 2H_2O + 2SO_2$
$2V 3V 2V$
128 g of $SO _2$ gives $=2 \times 22.4$ litres volume
So, 12.8 g of $SO _2$ gives $=2 \times 22.4 \times 12.8 / 128$
$=4.48$ litre volume
Or one can say 4.48 litres of hydrogen sulphide.
$222.4$ litre $H _2 S$ requires oxygen $=3 \times 22.4$ litre
So, $4.48$ litres $H _2 S$ will require $=6.72$ litre of oxygen
View full question & answer→Question 183 Marks
Solid ammonium dichromate decomposes as:
$
\left( NH _4\right)_2 Cr _2 O _7 \longrightarrow N _2+ Cr _2 O _3+4 H _2 O
$
If $63 g$ of ammonium dichromate decomposes. Calculate calculate the mass of chromium (III) oxide formed at the same time.
AnswerSolid ammonium dichromate decomposes as:
$
\left( NH _4\right)_2 Cr _2 O _7 \longrightarrow N _2+ Cr _2 O _3+4 H _2 O
$
$252 g$ of ammonium dichromate gives $152 g$ of $CrO _3$
$63 g$ of ammonium dichromate gives $\frac{63 \times 152}{252}$
$
=38 g
$
View full question & answer→Question 193 Marks
Solid ammonium dichromate decomposes as :
$
\left( NH _4\right)_2 Cr _2 O _7 \longrightarrow N _2+ Cr _2 O _3+4 H _2 O
$
If $63 g$ of ammonium dichromate decomposes. Calculate the volume of $N _2$ evolved at STP.
AnswerSolid ammonium dichromate decomposes as :
$
\left( NH _4\right)_2 Cr _2 O _7 \longrightarrow N _2+ Cr _2 O _3+4 H _2 O
$
$252 g$ of ammonium dichromate gives $22.4 dm ^3$ of $N _2$
$63 g$ of ammonium dichromate gives $\frac{63 \times 22.4}{252}$
$
=5.6 L
$
View full question & answer→Question 203 Marks
Solid ammonium dichromate decomposes as :
$
\left( NH _4\right)_2 Cr _2 O _7 \longrightarrow N _2+ Cr _2 O _3+4 H _2 O
$
If $63 g$ of ammonium dichromate decomposes. Calculate the quantity in moles of nitrogen formed
AnswerSolid ammonium dichromate decomposes as:
$
\left( NH _4\right)_2 Cr _2 O _7 \longrightarrow N _2+ Cr _2 O _3+4 H _2 O
$
$252 g$ of ammonium dichromate gives $22.4 dm ^3$ of $N _2$
$63 g$ of ammonium dichromate gives $\frac{63 \times 22.4}{252}$
$
=5.6 L
$
No. of moles $=\frac{\text { Weight }}{\text { Molecular weight }}$
$
=0.25 \text { moles }
$
View full question & answer→Question 213 Marks
Solid ammonium dichromate decomposes as :
$\left( NH _4\right)_2 Cr _2 O _7 \longrightarrow N _2+ Cr _2 O _3+4 H _2 O$
If $63 g$ of ammonium dichromate decomposes. Calculate the quantity in moles of $\left( NH _4\right)_2 Cr _2 O _7$
AnswerSolid ammonium dichromate decomposes as:
$\left( NH _4\right)_2 Cr _2 O _7 \longrightarrow N _2+ Cr _2 O _3+4 H _2 O$
Molecular mass of ammonium dichromate
$=2(14+4)+104+112 $
$=252 g$
$\text { Number of moles }=\frac{\text { Weight }}{\text { Molecular weight }} $
$=\frac{63}{252} $
$=0.25 \text { moles }$
View full question & answer→Question 223 Marks
What mass of copper hydroxide is precipitated by using $200 gm$ of sodium hydroxide?
$2 NaOH + CuSO _4 \longrightarrow Na _2 SO _4+ Cu ( OH )_2 \downarrow $
${[ Cu =64, Na =23, S =32, H =1]}$
Answer$2 NaOH + CuSO _4 \longrightarrow Na _2 SO _4+ Cu ( OH )_2 \downarrow $
$ \text { Molecular mass of } Cu ( OH )_2=64+16(2)+1(2)$
$ =98 $
$\text { Molecular mass of } NaOH =23+16+1=40$
$ \therefore 80 g \text { sodium hydroxide forms }=98 g \text { of } Cu ( OH )_2 $
$ \therefore 200 g \text { sodium hydroxide will from }=\frac{98 \times 200}{80}=245 g \text { of } Cu ( OH )_2$
View full question & answer→Question 233 Marks
Find the percentage of phosphorus in the fertilizer superphosphate $Ca(H_2PO_4)_2$
AnswerThe relative molecular mass of $Ca \left( H _2 PO _4\right)_2$
$=40+2(2+31+64)=234$
Since, $234 g$ of $Ca \left( H _2 PO _4\right)_2$ contains $62 g$ of phosphorus
$100 g \text { of } Ca \left( H _2 PO _4\right)_2 \text { contains } \frac{100 \times 62}{234}$
$=26.5 \%$
View full question & answer→Question 243 Marks
Find the percentage of boron in $Na_2B_4O_7.10H_2O. [H = 1, B = 11, O =16,$ Na $= 23].$
AnswerThe relative molecular mass of $Na _2 B _4 O _7 \cdot 10 H _2 O$
$(23 \times 2)+(4 \times 11)+(7 \times 16)+10(18)=382$
Since $382 g$ of $Na _2 B _4 O _7 .10 H _2 O$
contains $44 g$ of boron $100 g Na _2 B _4 O _7 .10 H _2 O$ of
contains $\frac{100 \times 44}{382}$ of boron $=11.5 \%$
View full question & answer→Question 253 Marks
Find the percentage of oxygen in magnesium nitrate crystals $[Mg (NO_3) 6H_2O]$.
AnswerThe relative molecular mass of $\left[ Mg \left( NO _3\right)_2 6 H _2 O \right]$
$=24+2(14+(3 \times 16))+(6 \times 18)=256$
Since, $256 g$ of $\left[ Mg \left( NO _3\right)_2 6 H _2 O \right]$ contains $192 g$ of oxygen
$\therefore 100 g$ of $\left[ Mg \left( NO _3\right)_2 6 H _2 O \right]$ contains $\frac{192 \times 100}{256}$ of oxygen $=75 \%$
View full question & answer→Question 263 Marks
From the equation :
$C +2 H _2 SO 4 \rightarrow CO _2+2 H _2 O +2 SO _2$
Calculate :
1. The mass of carbon oxidized by $49 g$ of sulphuric acid.
2. The volume of sulphur dioxide measured at $STP$, liberated at the same time.
Answer$C + 2H_2SO4 \rightarrow CO_2 + 2H_2O + 2SO_2$
For $CO_212+32$
i. Molecular mass of sulphuric acid $= 2(2+32+64)$
$= 196$
196 g of suphuric acid oxidized $12g$ of Carbon $49 g$ of suphuric acid will $`(12 xx 49)/196`$
$= 3 g$
ii. $196 g$ of sulphuric acid gives 2(22.4)
$= 44.8L$
$\therefore 49 g$ of sulphuric acid will give $`(44.8 xx 49)/196`$
$= 11.2 L$ of $SO_2$
View full question & answer→Question 273 Marks
What would be the mass of $CO_2$ occupying a volume of $44$ litres at $25^0 C$ and $750$ mm pressure?
Answer$ P _1=750 mm , P _2=760 mm$
$V _1=44 lit . V _2=?$
$T _1=298 K T _2=273 K$
$\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}$
$V_2=\frac{P_1 V_1 T_2}{T_1 P_2}=\frac{750 \times 44 \times 273}{298 \times 760}=39.78 \text { lit. } $
$22.4$ lit. of $CO _2$ at STP has mass $=44 g$
$39.78$ lit. of $CO _2$ at STP has mass $=\frac{44 \times 39.78}{22.4}$ $=78.14 g$
View full question & answer→Question 283 Marks
How much volume will be occupied by $2g$ of dry oxygen at $27^0C$ and $740$ mm pressure?
Answer$32 g$ of dry oxygen at STP $=22400 cc$
2 g will occupy $=224002 / 32=1400$ cc
$P_1=760 mm P_2=740 mm$
$V _1=1400 cc V _2=?$
$T _1=273 K , T _2=27+73=300 K$
$\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}$
$V_2=\frac{P_1 V_1 T_2}{T_1 P_2}=\frac{760 \times 1400 \times 300}{273 \times 740}=1580 cc$
$=1580 / 1000=1.581$
View full question & answer→Question 293 Marks
What volume of hydrogen and oxygen measured at S.T.P. will be required to prepare 1.8 g of water?
AnswerFor $18 g$ water, vol. of hydrogen needed $=22.4$ litre
So, for $1.8 g$, vol. of $H _2$ needed $=1.8 \times 22.4 / 18=2.24$ litre
Now 2 vols. of water = 1 vol. of oxygen
1 vol. of water $=1 / 2$ vol. of $O _2=22.4 / 2=11.2$ lit.
$18 g$ of water $=11.2$ lit. of $O _2$
$1.8 g$ of water $=11.2 / 18 \Rightarrow 18 / 10=1.12$ lit.
View full question & answer→Question 303 Marks
The following questions refer to one mole of chlorine gas.
(a) What is the volume occupied by this gas at S.T.P.?
(b) What will happen to the volume of gas, if pressure is doubled?
(c) What volume will it occupy at $273^{\circ} C$ ?
(d) If the relative atomic mass of chlorine is 35.5, what will be the mass of 1 mole of chlorine gas?
Answer(a) The volume occupied by 1 mole of chlorine $=22.4$ litre
(b) Since PV=constant so, if pressure is doubled; the volume will become half i.e. 11.2 litres.
(c) $V_1 / V_2=T_1 / T_2$
$22.4 / V_2=273 / 546$
$V _2=44.8$ litres
(d) Mass of 1 mole $Cl _2$ gas $=35.5 \times 2=71 g$
View full question & answer→Question 313 Marks
A gas cylinder can hold $1 kg$ of hydrogen at room temperature and pressure.
(a) What mass of carbon dioxide can it hold under similar conditions of temperature and pressure?
(b) If the number of molecules of hydrogen in the cylinder is $X$, calculate the number of carbon dioxide molecules in the cylinder. State the law that helped you to arrive at the above result
Answer(a) Molecular mass of $CO _2=12+2 \times 16=44 g$
So, vapour density (V.D) $=$ mol. Mass $/ 2=44 / 2=22$
$ V . D=\frac{\text { mass of certain amount of } CO _2}{\text { mass of equal volume of hydrogen }}=\frac{ m }{1}$
$22=\frac{ m }{1} $
So, mass of $CO _2=22 kg$
(b) According to Avogadros law, equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.
So, number of molecules of carbon dioxide in the cylinder = number of molecules of hydrogen in the cylinder=X
View full question & answer→Question 323 Marks
Silicon (Si = 28) forms a compound with chlorine (Cl = 35.5) in which 5.6 g of silicon combines with 21.3 g of chlorine. Calculate the empirical formula of the compound.
AnswerMass of silicon in the given compound $=5.6 g$
Mass of the chlorine in the given compound $=21.3 g$
The total mass of the compound $=5.6 g +21.3 g =26.9 g$
$\%$ of silicon in the compound $=56 / 26.9 \times 100=20.82 \%$
$\%$ of chlorine in the compound $=21.2 / 26.9 \times 100=79.18 \%$
Element \% At. Mass At. Ratio Simplest ratio
Si $20.82: 28: 20.82 / 28=0.74: 1$
Cl $79.18: 35.5: 79.18 / 35.5=2.23: 3$
So the empirical formula of the given compound $= SiCl _3$
View full question & answer→Question 333 Marks
The equation for the burning of octane is:
$2 C _8 H _{18}+25 O _2 \rightarrow 16 CO _2+18 H _2 O$
(i) How many moles of carbon dioxide are produced when one mole of octane burns?
(ii) What volume at S.T.P. is occupied by the number of moles determined in (i)?
(iii) If the relative molecular mass of carbon dioxide is 44 , what is the mass of carbon dioxide produced by burning two moles of octane?
(iv) What is the empirical formula of octane?
Answer$\ce{2C8H18 + 25O2 -> 16CO2 + 18H2O}$
2 V25 V16 V18 V
(i) 2 moles of octane gives $=16$ moles of $CO _2$
so, 1 mole octane will give $=8$ moles of $CO _2$
(ii) 1 mole $CO _2$ occupies volume $=22.4$ litre
so, 8 moles will occupy volume $=822.4=179.2$ litre
(iii) 1 mole $CO _2$ has mass $=44 g$
so, 16 moles will have mass $=4416=704 g$
(iv) Empirical formula is $C _4 H _9 . ff$
View full question & answer→Question 343 Marks
How much calcium oxide is obtained by heating $82 g$ of calcium nitrate? Also, find the volume of $NO _2$ evolved:
$
2 Ca \left( NO _3\right)_2 \longrightarrow 2 CaO +4 NO _2+ O _2
$
Answer$
2 Ca \left( NO _3\right)_2 \longrightarrow 2 CaO +4 NO _2+ O _2
$
Molecular Weight of $2 Ca ( NO ) 2=2[40+2(14+48)]=328 g$
Molecular weight of $CaO =2(40+16)=112 g$
a. $328 g$ of $Ca \left( NO _3\right)_2$ liberates 4 moles of $NO _2$
$328 g$ of $Ca \left( NO _3\right)_2$ liberates $4 \times 22.4 L$ of $NO _2$
$\therefore 82 g$ will liberate $\frac{4 \times 22.4}{328} \times 82=22.4 dm ^3$ of $NO _2$
b. $328 g$ of calcium nitrate gives $112 g$ of $CaO$
$82 g$ will give $\frac{112 \times 82}{328}$
$
=28 g \text { of } CaO
$
View full question & answer→Question 353 Marks
$60 cm ^3$ of oxygen was added to $24 cm ^3$ of carbon monoxide and mixture ignited. Calculate:
(i) The volume of oxygen used up and
(ii) The volume of carbon dioxide formed.
Answer$2 CO + O _2 \rightarrow 2 CO _2$
$2 V 1 V 2 V$
(i) $2 V CO$ requires oxygen $=1 V$
So, $24 cm ^3 CO$ will require $=24 / 2=12 cm ^3$
(ii) $2 \times 22400 cm ^3 CO$ gives $=2 \times 22400 cm ^3 CO _2$
so, $24 cm ^3 CO$ will give $=24 cm ^3 CO _2$
View full question & answer→Question 363 Marks
(a) Calculate the mass of ammonia that can be obtained from $21.4 g$ of $NH _4 Cl$ by the reaction:
$2 NH _4 Cl + Ca ( OH )_2 \rightarrow CaCl _2+2 H _2 O +2 NH _3$
(b) What will be the volume of ammonia when measured at S.T.P?
The molar volume of a gas $=22.4$ litres at STP.
Answer$ 2 NH _4 Cl + Ca ( OH )_2 CaCl _2+2 H _2 O +2 NH _3$
$2 V 1 V 1 V 2 V $
Mol. Mass of $2 NH _4 Cl =2[14+(1 \times 4)+35.5]=2[53.5]=107 g$
(a) $107 g NH _4 Cl$ gives $=34 g NH _3$
So, $21.4 g NH _4 Cl$ will give $=21.4 \times 34 / 107=6.8 g NH _3$
(b) The volume of $17 g NH _3$ is $22.4$ litre
So, volume of $6.8 g$ will be $=6.8 \times 22.4 / 17=8.96$ litre
View full question & answer→Question 373 Marks
Calculate the mass and volume of oxygen at S.T.P., which will be evolved on electrolysis of 1 mole (18g) of water
Answer$
2 H _2 O \rightarrow 2 H _2+ O _2
$
$2 V 2 V 1 V$
2 moles of $H _2 O$ gives $=1$ mole of $O _2$
So, 1 mole of $H _2 O$ will give $=0.5$ moles of $O _2$
so, mass of $O _2=$ no. of moles $x$ molecular mass
$
=0.5 \times 32=16 g \text { of } O _2
$
and 1 mole of $O _2$ occupies volume $=22.4$ litre
so, 0.5 moles will occupy $=22.4 \times 0.5=11.2$ litres at S.T.P.
View full question & answer→Question 383 Marks
Find the mass of $KNO _3$ required to produce $126 kg$ of nitric acid. Find whether a larger or smaller mass of $NaNO _3$ is required for the same purpose.
$
KNO _3+ H _2 SO _4 \rightarrow KHSO _4+ HNO _3
$
$
NaNO _3+ H _2 SO _4 \rightarrow NaHSO _4+ HNO _3
$
AnswerThe molecular mass of $KNO _3=101 g$
$63 g$ of $HNO _3$ is formed by $=101 g$ of $KNO _3$
So, $126000 g$ of $HNO _3$ is formed by $=126000 \times 101 / 63=202 kg$
Similarly, $126 g$ of $HNO _3$ is formed by $170 kg$ of $NaNO _3$
So, a smaller mass of $NaNO _3$ is required.
View full question & answer→Question 393 Marks
$66 g$ of ammounium sulphate is produced by the action of ammonia on sulphuric acid.
Write a balanced equation and calculate:
(a) Mass of ammonia required.
(b) The volume of the gas used at the S.T.P.
(c) The mass of acid required.
Answer$
2 NH _3+ H _2 SO _4\left( NH _4\right)_2 SO _4
$
$
66 g
$
(a) $2 NH _3+ H _2 SO _4\left( NH _4\right)_2 SO _4$
$
34 g 98 g 132 g
$
For $132 g \left( NH _4\right)_2 SO _4=34 g$ of $NH _3$ is required
So, for $66 g \left( NH _4\right)_2 SO _4=66 \times 32 / 132=17 g$ of $NH _3$ is required
(b) $17 g$ of $NH _3$ requires volume $=22.4$ litres
(c) Mass of acid required, for producing $132 g \left( NH _4\right)_2 SO _4=98 g$
So, Mass of acid required, for $66 g \left( NH _4\right)_2 SO _4=66 \times 98 / 132=49 g$
View full question & answer→Question 403 Marks
Nitrogen and hydrogen react to form ammonia.
$N _2( g )+3 H _2( g ) 2 NH _3( g )$
If $1000 g H _2$ reacts with $2000 g$ of $N _2$ :
(a) Will any of the two reactants remain unreacted? If yes, which one and what will be its mass?
(b) Calculate the mass of ammonia( $\left( NH _3\right)$ that will be formed?
Answer$ N _2+3 H _2 2 NH _3$
$28\ g\ 6\ g\ 34\ g $
$28 g$ of nitrogen requires hydrogen $=6 g$
$2000 g$ of nitrogen requires hydrogen $=6 / 28 \times 2000=3000 / 7 g$
So mass of hydrogen left unreacted $=1000-3000 / 7=571.4 g$ of $H _2$
(b) $28 g$ of nitrogen forms $NH _3=34 g$
$2000 g$ of $N _2$ forms $NH _3$
$=34 / 28 \times 2000$
$=2428.6 g$
View full question & answer→Question 413 Marks
$
MnO _2+4 HCl \rightarrow MnCl _2+2 H _2 O + Cl _2
$
0.02 moles of pure $MnO _2$ is heated strongly with conc. $HCl$. Calculate: Mass of acid required
Answer$
MnO _2+4 HCl \rightarrow MnCl _2+2 H _2 O + Cl _2
$
1 V4 V1 V1 V
For 1 mole $MnO _2$, acid required $=4$ mole of $HCl$
So, for 0.02 mole, acid required $=4 \times 0.02=0.08$ mole
$
\text { Mass of } HCl =0.08 \times 36.5=2.92 g
$
View full question & answer→Question 423 Marks
Aluminium carbide reacts with water according to the following equation.
$
Al _4 C _3+12 H _2 O \rightarrow 3 CH _4+4 Al ( OH )_3
$
(a) What mass of aluminium hydroxide is formed from $12 g$ of aluminium carbide?
(b) What volume of methane s.t.p. is obtained from $12 g$ of aluminium carbide?
Answer$
\underset{1 V , 144 g }{ Al _4 C _3}+12 H _2 O \longrightarrow \underset{3 V , 3 \times 22.4 \text { vol }}{3 CH _4}+\underset{312 g }{4 Al ( OH )_3}
$
(i)
Since $144 g$ of $Al _4 C _3$ gives $312 g$ of $Al ( OH )_3$
So, $12 g$ of $Al _4 C _3$ will give $=\frac{312 \times 12}{144}$
$
=26 g Al ( OH )_3
$
(ii) Now, Since $144 g$ of $Al _4 C _3$ gives $3 \times 22.4$ litre of $CH _4$
So, $12 g$ of $Al _4 C _3$ will give $=\frac{3 \times 22.4 \times 12}{144}$
$
=5.6 \text { litre } CH _4
$
View full question & answer→Question 433 Marks
Calculate the mass of iron in 10 kg of iron ore which contains 80% of pure ferric oxide.
Answer$
\text { In } Fe _2 O _3, Fe =56 \text { and } O =16
$
M0olecular mass of $Fe _2 O _3=2 \times 56+3 \times 16=160 g$
Iron present in $80 \%$ of $Fe _2 O _3=\frac{112}{160} \times 80=56 g$
So, mass of iron in $100 g$ of ore $=56 g$
mass of $Fe$ in $10000 g$ of ore $=56 \times 10000 / 100$
$
=5.6 kg
$
View full question & answer→Question 443 Marks
Find the empirical formula of the compounds with the following percentage compositi
Find the empirical formula of the compounds with the following percentage composition:
$
P b=62.5 \%, N=8.5 \%, O=29.0 \%
$
on:
$
Pb =62.5 \%, N =8.5 \%, O =29.0 \%
$
AnswerElement $\%$ At. mass Atomic ratio Simple ratio Pb $62.5207 \frac{62.5}{207}=0.3019_1$
$N 8.514 \frac{8.5}{14}=0.6071_2$
$29.016 \frac{29.0}{16}=1.81_6$
$So , Pb \left( NO _3\right)_2$ is the empirical formula.
View full question & answer→Question 453 Marks
Find the percentage of water of crystallisation in $CuSO_4.5H_2O$. (At. Mass $Cu = 64, H = 1, O = 16, S = 32)$
AnswerThe relative molecular mass of $CuSO _4 \cdot 5 H _2 O$
$ =64+32+4 \times 16+5(2+16)$
$=160+90$
$=250 $
$250 g$ of $CuSO _4 \cdot 5 H _2 O$ contains $90 g$ of water of crystallization.
$ =\frac{90}{250} \times 100$
$=36 \% $
View full question & answer→Question 463 Marks
A compound has O=61.32%, S= 11.15%, H=4.88% and Zn=22.65%.The relative molecular mass of the compound is 287 amu. Find the molecular formula of the compound, assuming that all the hydrogen is present as a water of crystallization.
AnswerElement $\%$ at. mass atomic ratio simple ratio
Zn 22.65650 .3481
H 4.8814 .8814
S 11.15320 .3481
O 61.32163 .8311
Empirical formula of the given compound $= ZnSH _{14} O _{11}$
Empiricala formula mass $=65.37+32+141+11+16=287.37$
Molecular mass $=287$
$
n =\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{287}{287}=1
$
Molecular formula $= ZnSO _{11} H _{14}$
$
= ZnSO _4 \cdot 7 H _2 O
$
View full question & answer→Question 473 Marks
Give three kinds of information conveyed by the formula $H_2O.$
AnswerInformation conveyed by $H _2 O$
(1)That $H _2 O$ contains 2 volumes of hydrogen and 1 volume of oxygen.
(2)That ratio by weight of hydrogen and oxygen is 1:8.
(3)The molecular weight of $H _2 O$ is $18 g$.
View full question & answer→Question 483 Marks
Barium chloride crystals contain $14.8\%$ water of crystallization. Find the number of molecules of water of crystallization per molecule.
Answer$\text { Barium chloride }= BaCl _2 \cdot H _2 O $
$ Ba +2 Cl +x\left[ H _2+ O \right] $
$ =137+235.5+x[2+16] $
$ =[208+18 x] \text { contains water }=14.8 \% \text { water in } BaCl _2 \cdot H _2 O$
$=[208+18 x] 14.8 / 100=18 x $
$ =[104+9 x] 2148=18000 x$
$ =[104+9 x] 37=250 x $
$ =3848+333 x=2250 x$
$ 1917 x=3848 $
$ x=2 \text { molecules of water }$
View full question & answer→Question 493 Marks
0.2 g atom of silicon Combine with 21.3 g of chlorine. Find the empirical formula of the compound formed.
AnswerSince $0.2 g$ atom of $Si =$ given mass $/ mol$. Mass
so, given mass $=0.2 \times 28=5.6 g$
Element mass $/$ At. mass $=$ Gram atom Ratio
Si $5.6 / 28=0.21$
Cl 21.3/35.5 $=0.63$
$
\text { Si:Cl }=0.21: 0.63=1: 3
$
now we see that one part of silicon reacts with 3 parts of chlorine so, the empirical formula is $SiCl _3$
View full question & answer→Question 503 Marks
An organic compound contains $H =4.07 \%, Cl =71.65 \%$ chlorine and remaining carbon. Its molar mass $= 98.96.$ Find,
(a) Empirical formula, and
(b) Molecular formula
AnswerElement $\%$ at. mass atomic ratio simple ratio
$\text { Cl } 71.6535 .5 \frac{71.65}{35.5}=2.011$
$\text { H } 4.071 \frac{4.07}{1}=4.07_2 $
$\text { C } 24.2812 \frac{24.28}{12}=2.02_1$
(a) its empirical formula $= CH _2 Cl$
(b) empirical formula mass $=49.5$
Since, molecular mass $=98.96$
so, molecular formula $=\left( CH _2 Cl \right)_2= C _2 H _4 Cl _2$
View full question & answer→