MCQ

MCQ 11 Mark

Which of the following contains maximum number of molecules?

A
4 g of $O _2$
✓
4 g of $NH _3$
C
4 g of $CO _2$
D
4 g of $SO _2$

Answer

Correct option: B.

4 g of $NH _3$

4 g of $NH _3$ having minimum molecular mass contain maximum molecules.

View full question & answer→

MCQ 21 Mark

The number of molecules in 4.25 g of ammonia is :

A
$1.0 x 10^{23}$
✓
$1.5 x 10^{23}$
C
$2.0 x 10^{23}$
D
$3.5 x 10^{23}$

Answer

Correct option: B.

$1.5 x 10^{23}$

The number of molecules in 18 g of ammonia $=6.02 \times 10^{23}$ So, no. of molecules in 4.25 g of ammonia $=6.02 \times 10^{23} \times 4.25 / 18$ $=1.5 \times 10^{23}$

View full question & answer→

MCQ 31 Mark

Four grams of caustic soda contains :

A
$6.02 \times 10^{23}$ atoms of it
B
4 g atom of sodium
✓
$6.02 \times 10^{22}$ molecules
D
4 moles of NaOH

Answer

Correct option: C.

$6.02 \times 10^{22}$ molecules

Option C is correct.
40 g of NaOH contains $6.023 \times 10^{23}$ molecules
So, 4 g of NaOH contains $=6.02 \times 10^{23} \times 4 / 40$
$=6.02 \times 10^{22}$ molecules

View full question & answer→

MCQ 41 Mark

Assertion: The reactant which is present in lesser amount parameters the amount of product formed is called limiting reagent.
Reason: Amount of product formed does not depend upon the number of reactants taken.

A
Assertion and Reason both are correct statements and reason is the correct explanation of the assertion
B
Assertion and Reason both are correct statements, but reason is not the correct explanation of the assertion
C
Assertion is true, but reason is false.
D
Assertion is false, but reason is true.

Answer

C. Assertion is true, but reason is false.
Explanation: The reactant which is present in lesser amount gets consumed after some time and after that no further reaction takes place whatever be the amount of the other reactant present. Hence, the reactant which gets consumed, limits the amount of product formed and is, therefore, called the limiting reagent. Thus, assertion is true, but reason is false.

View full question & answer→

MCQ 51 Mark

The simplest formula of a compound containing $50 \%$ of element $X$ (atomic mass 10 ) and $50 \%$ of element $Y$ (atomic mass 20) is:

A
$X Y$
B
$X _2 Y$
C
$X Y_3$
D
$X _2 Y _3$

Answer

(b) $X_2 Y$
Explanation:

Element	Mass \%	Atomic mass	Relative no. of moles	Simplest Ratio
X	50	10	$\frac{50}{10}=5$	2
Y	50	20	$\frac{50}{20}=2.5$	1

Therefore the simplest formula (empirical formula) $= X _2 Y$.

View full question & answer→

MCQ 61 Mark

If two compounds have the same empirical formula but different molecular formula, they must have

A
Different percentage composition
B
Different molecular weights
C
Same viscosity
D
Same vapour density

Answer

B. Different molecular weights
Explanation: If two compounds have the same empirical formula but different molecular formula, they must have different molecular weights. For example, CH₂O and C₆H₁₂O₆ have the same empirical formula but different molecular formula, they have different molecular weights.

View full question & answer→

MCQ 71 Mark

The percentage composition of carbon in urea, $\left[ CO \left( NH _2\right)_2\right]$ is :

A
$40 \%$
B
$50 \%$
C
$20 \%$
D
$80 \%$

Answer

C. 20\%
Explanation:
Molar mass of $\left[ CO \left( NH _2\right)_2\right]$ is $12+16+2(14+2)=60 g / mol$
$\text { Mass \% of an element }=\frac{\text { Mass of that element in the compound }}{\text { Molar mass of the compound }} \times 100$
Percentage composition of Carbon $=\frac{12}{60} \times 100=20 \%$

View full question & answer→

MCQ 81 Mark

4.4 g of an unknown gas occupies 2.24 L of volume at standard temperature and pressure. The gas may be:

A
Carbon dioxide
B
Carbon monoxide
C
Oxygen
D
Sulphur dioxide

Answer

A. Carbon dioxide
Explanation : According to mole concept, 1 mole of any substance = molar mass = 6.02 $\times$ 10²³ particles
= 22.4 L volume
If, 2.24 L of a gas weight 4.4 g
Then 22.4 L of a gas will weigh 44 g
It is the molar mass of carbon dioxide.

View full question & answer→

MCQ 91 Mark

The weight of $1 \times 10^{22}$ molecules of $CuSO _4 \cdot 5 H _2 O$ is :

A
41.59 g
B
415.9 g
C
4.159 g
D
None of these

Answer

C. 4.159 g
Explanation:
Molar mass of $CuSO _4 \cdot 5 H _2 O =63.5+32+4(16)+5(2+16)$
$=249.5 g / mol$
1 mole that is $6.023 \times 10^{23}$ molecules of $CuSO _4 \cdot 5 H _2 O$ has 249.5 g of mass.
Then, $1 \times 10^{22}$ molecules of $CuSO _4 \cdot 5 H _2 O$ will have $=\frac{249.5 \times 1 \times 10^{22}}{6.023 \times 10^{23}}=4.158 g$

View full question & answer→

MCQ 101 Mark

Total number of atoms represented by the compound CuSO₄.5H₂O is:

A
27
B
21
C
5
D
8

Answer

B. 21
Explanation: The total number of atoms of each element that combines to form a compound is known as its atomicity. In the given compound, number of atoms of Cu is 1, S is 1, O is (4 + 5) = 9, H is 5(2) = 10. Hence, total number of atoms=1+1+9+10=21

View full question & answer→

MCQ 111 Mark

Avogadro number is:

A
Number of atoms in one gram of element
B
Number of milliliters which one mole of a gaseous substances occupies at NTP
C
Number of molecules present in one gram molecular mass of a substance
D
All of the above

Answer

C. Number of molecules present in one gram molecular mass of a substance
Explanation: The number of entities present in 1 mol (gram molecular mass) of any substance is represented as Avogadro's constant or Avogadro's number denoted by N_A. One mole contains exactly 6.02214076 × 10²³ elementary entities.

View full question & answer→

MCQ 121 Mark

What is the mass of 3 moles of carbon dioxide $\left( CO _2\right)$ ? (The molar mass of $CO _2$ is approximately $44 g / mol$ )

A
44 g
B
88 g
C
132 g
D
176 g

Answer

C. 132 g
Explanation: The molar mass of carbon dioxide $\left( CO _2\right)$ is $44 g / mol$. The mass of 1 mole of $CO _2$ is 44 grams. Therefore, the mass of 3 moles would be 3 times 44 g , which is equal to 132 g .

View full question & answer→

MCQ 131 Mark

The molecular formula of the china clay (aluminium silicate) was shown in old book as $Al _2 O _3 \cdot 2 SiO _2 \cdot 2 H _2 O$.
The formula is shown in modern book as $Al _2 \cdot Si _2 O _y \cdot( OH )_x$
What are valencies of $x$ and $y$ in the modern formula?

A
y x
2 4
B
y x
2 5
C
y x
4 3
D
y x
4 5

Answer

D. $y=4, x=5$
Explanation:

	$Al _2 O _3 \cdot 2 SiO _2 \cdot 2 H _2 O$	$A _{12} \cdot Si _2 O y \cdot( OH ) x$
Al atoms	2	2
Si atoms	2	2
O atoms	$3+2(2)+2(1)=9$	$y+x$
H atoms	2(2) = 4	y

On comparing Hydrogen atoms $y=4$
Now comparing oxygen atoms,
$\begin{array}{l}
9=y+x \\
9=4+x \\
x=5
\end{array}$

View full question & answer→

MCQ 141 Mark

Calcium reacts with water as shown.
$Ca(s)+H_2 O_{(l)} \square \rightarrow Ca(OH)_{2(a q)}+H_{2(g)}$
What is the total mass of the solution that remains when 40 g of Calcium reacts with 100 g of water.

A
58 g
B
74 g
C
138 g
D
140 g

Answer

D. 140 g
Explanation:
Assume that the reaction goes to completion.
$Ca_{(s)}+H_2 O_{(l)} \rightarrow Ca(OH)_{2(a q)}+H_{2(g)}$
40 g of $Ca =40 g(40 g$ mole $)-1$ mole of Ca
Mol. ratio of $Ca : H _2 O =1: 2$
Thus, 1 mol of Ca reacts with $2 \times 1 mol$ of $H _2 O -2 mol$ of $H _2 O -2 mol-36 g$ of $H _2 O$
Since, the available $H _2 O =100 g$, mass of Cs is 40 g , and mass of $H _2 O$ used up 36 g , so the limiting reactant is Ca.
Then, then $\qquad$ $H _2 O =(100 g-36 g)=68 g$.
Mol. ratio of $Ca : Ca ( OH )_2=1: 1$
Mol ratio of $Ca : H _2=1: 1$
Thus, I and of Cs produces I and of $Ca ( OH )_2$ and I and of $H _2$.
Mass of $Ca ( OH )_2=1 mol \times 74 g mol =74 g$
Mass of $H _2=1 mol \times 74 g mol =74 g$
Mass of $H _2=1 mol \times 2 g mol =2 g$
Then, total mass of the solutions after reactions $=[$ mass of $Ca ( OH ) \times 2+64 g+74 g+2 g=14 g$

View full question & answer→

MCQ 151 Mark

In a sample of NO:
P: Percentage by mass of Nitrogen - $47 \%$
Q: Percentage by mass of Oxygen - $53\%$
Which statements are correct?

A
Only P
B
Only Q
C
Both P and Q
D
None of these

Answer

C. Both P and Q
Explanation:

	Percentage	Molar mass	Moles	Molar ratio
N	47	14	3.35	1
O	53	16	3.31	1

Thus, the empirical formula of the compound will be NO.

View full question & answer→

MCQ 161 Mark

The percentage of oxygen $( O )$ in sulphur dioxide $\left[ SO _2\right]$ :

A
2.5
B
50
C
60
D
40

Answer

B. 50
Explanation:
Atomic mass of oxygen $=16$
Atomic mass of sulphur $=32$
Molecular mass of $SO _2=32+2(16)=64$
Mass of $O _2$ molecule $=16 \times 2=32$
$\%$ of [O] in sulphur dioxide $=\frac{32}{64} \times 100 \%=50 \%$

View full question & answer→

MCQ 171 Mark

In $Na _2 CO _3$, the percentage mass of oxygen is:

A
62.93
B
45.3
C
59.6
D
40.3

Answer

B. 45.3
Explanation:
The molecular mass of sodium carbonate is 106.
$(2 Na-46,1 C=12 \text {, and } 3 O=48)$
Since, the atomic mass of 3 oxygen atoms is 48
Therefore, $100 \times \frac{48}{106}=45.3 \%$

View full question & answer→

MCQ 181 Mark

Calculate the value of $x$, when the hydrated salt $Na _2 CO _3 \cdot x H _2 O$ undergoes $63 \%$ loss in mass on heating and becomes anhydrous.

A
3
B
5
C
7
D
10

Answer

D. 10
Explanation:
Molecular mass of $Na _2 CO _3=2 \times 23+12+16 \times 3=106 g$
To know the mass of $Na _2 CO _3$ containing $63 \% H _2 O$, we subtract 63 from 100
$\begin{array}{l}
100-63=37 g \\
37 g \text { of } Na_2 CO_3 \text { contains }=63 g H_2 O \\
1 g \text { of } Na_2 CO_3 \text { contains }=\frac{63}{37} g \text { of } H_2 O \\
106 g \text { of } Na_2 CO_3 \text { will contain }=\frac{63}{37} \times 106 \\
=180.48 g H_2 O
\end{array}$
Number of water molecules $x=\frac{180.48}{18} g\left(\frac{ wt }{ mol wt }\right)$
$x=10$

View full question & answer→

MCQ 191 Mark

If vapour density of the gas is 39 and has molecular formula $( CH )_{ n }$. Then what should be the formula of the compound?

A
$C _3 H _3$
B
$C _4 H _4$
C
$C _2 H _2$
D
$C _6 H _6$

Answer

D. $C _6 H _6$
Explanation:
Given that,
Molecular weight $=2 \times$ Vapour density
$=2 \times 39=78$
$(CH)_n=(13)_n$
or, $13 \times n=78$
$n=\frac{78}{13}=6$
Molecular formula $= C _6 H _6$

View full question & answer→

MCQ 201 Mark

The weight of lime obtained by heating 200 kg of $95 \%$ pure lime stone is:

A
98.4 kg
B
106.4 kg
C
112.8 kg
D
122.6 kg

Answer

B. 106.4 kg
Explanation:
Amount of pure $CaCO _3=\frac{95}{100} \times 200=190 kg=190000 g$
$CaCO _3 \xrightarrow{\Delta} CaO + CO _2$
$(40+12+3 \times 16)(40+16)$
10056
190 W
Since, 100 g of $CaCO _3$ on heating gives 56 g of lime.
$\therefore 190000 g CaCO _3$ will give ${ }_{100}^{56} \times 190000 g=106400 g=106.4 kg$
Hence, the weight of lime obtained by heating 200 kg of $95 \%$ pure lime stone is 106.4 Kg .

View full question & answer→

MCQ 211 Mark

If the empirical formula of a compound is CH and its vapour density is 13 . Its molecular formula will be:( $C =12, H =1$ )

A
CH
B
$C _2 H _2$
C
$C _4 H _4$
D
$C _3 H _3$

Answer

B. $C _2 H _2$
Explanation:
Given, empirical formula is CH .
Vapour density $=13$
Now, Molecular mass $=2 \times$ Vapour density
$=2 \times 13=26$
Empirical weight of $CH =12+1=13$
Now, $n=\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{26}{13}=2$
Molecular formula $=(\text { Empirical formula })_n$
$=(CH)_n=C_2 H_2$
Hence, its molecular formula will be $C _2 H _2$.

View full question & answer→

MCQ 221 Mark

What is the percentage of oxygen in ammonium nitrate ? (Nitrogen $=14$, Hydrogen $=1$, Oxygen $=$ 16).

A
$20 \%$
B
$30 \%$
C
$60 \%$
D
$80 \%$

Answer

C. 60%
Explanation:
Ammonium nitrate $= NH _4 NO _3$
Relative $mol . wt .=14+1 \times 4+14+3 \times 16=80$
Total nitrogen in ammonium nitrate $=14+14=28$
Percentage of nitrogen $=\frac{28}{80} \times 100=35 \%$
Total oxygen in ammonium nitrate $=16 \times 3=48$
Percentage of Oxygen $=\frac{48}{80} \times 10=60 \%$
Hence, the percentage of oxygen in ammonium nitrate is $60 \%$.

View full question & answer→

MCQ 231 Mark

Calculate the atomicity of a gas whose vapour density is equal to its relative molecular mass.

Answer

B. 2
Explanation:
Given,
Vapour density = relative atomic mass $=35.5$
Molecular weight $=2 \times$ vapour density $=2 \times 35.5=71 g$
Atomicity is the number of atoms present in one molecule of the element, Number of atoms
$=\frac{\text { Molecular weight }}{\text { Atomic weight }}=\frac{71}{355}=2$
Hence, atomicity of a gas $\times$ is 2 .

View full question & answer→

MCQ 241 Mark

Calculate the percentage of nitrogen in aluminium nitride. $( Al =27, N=14)$

A
$34.15 \%$
B
$23.27 \%$
C
$12.8 \%$
D
$25.6 \%$

Answer

A. $34.15 \%$
Explanation:
Step 1 : Atomic weight of nitrogen $( N )=14 g / mol$
Atomic weight of aluminium $( Al )=27 g / mol$.
$\text { Step 2: Mass percentage }=\frac{\text { Atomic weight }}{\text { Total molecular weight }} \times 100$
Step 3 : Molecular weight of aluminum nitride (AIN)
$=27+14=41 g / mol$
Mass percentage composition of nitrogen
$=\frac{14}{41} \times 100=34.15 \%$
Therefore, mass percentage of nitrogen in aluminium nitride is $34.15 \%$

View full question & answer→

MCQ 251 Mark

What is the percentage mass of copper in blue vitriol crystal?

A
$25.45 \%$
B
$36.07 \%$
C
$49.56 \%$
D
None of these

Answer

A. $25.45 \%$
Explanation:
Blue vitriol is copper sulphate pentahydrate $CuSO _4 \cdot 5 H _2 O$
The molecular weight of $CuSO _4 \cdot 5 H _2 O$ is:
$63.5+32+4(16)+5(18)=249.5 g / mol$
The percentage of copper in sample of blue vitriol is,
$\frac{63.5 \times 100}{249.5}=25.45 \%$

View full question & answer→

MCQ 261 Mark

A compound with empirical formula AB has vapour density three times its empirical formula. Its molecular formula will be:

A
A₆B₆
B
A₂B₄
C
A₂B₄
D
AB

Answer

A. $A _6 B_6$
Explanation:
Molecular formula $=($ Empirical formula $) \times n$
$n=\frac{\text { Molecular mass }}{\text { Empirical mass }} \ldots \text { (i) }$
Vapour density $=\frac{\text { Molecular mass }}{2}$
Molecular mass $=2 \times$ Vapour density. $\ldots(ii)$
From question: Vapour density $=3 \times$ Empirical Mass...(iii)
From (ii) and (iii),
Molecular mass $=2 \times 3 \times$ Empirical mass
Molecular mass $=6 \times$ Empirical mass
$6=\frac{\text { Molecular mass }}{\text { Empirical mass }} \ldots \text { (iv) }$
From (i) and (iv),
Molecular formula $=($ Empirical formula $) \times n$
$\begin{array}{l}
=(AB) \times n \\
=(AB) \times 6
\end{array}$
Molecular formula $= A _6 B_6$.

View full question & answer→

MCQ 271 Mark

A compound with empirical formula AB₂ has the vapour density equal to its empirical formula weight. Its molecular formula is:

A
A₂B₂
B
A₂B₄
C
A₂B₃
D
A₂B₈

Answer

B. $A _2 B_4$
Explanation :
Given,
Empirical formula $= AB _2$
Vapour density $=\frac{\text { Molecular weight }}{2}$
Molecular weight $=2 \times VD$
According to the question,
Vapour density $=$ Empirical formula weight
Molecular formula $=($ Empirical formula $) n$
$\begin{array}{l}
n=\frac{\text { Molecular weight }}{\text { Empirical formula weight }} \\
=\frac{2 \times VD}{VD} \\
n=2
\end{array}$
So, molecular formula $=\left( AB _2\right)_2$
$=A_2 B_4$
Hence, the molecular formula is $A _2 B_4$.

View full question & answer→

MCQ 281 Mark

The molecular formula of a gas with vapour density 15 and empirical formula CH₃ is:

A
C₂H₆
B
C₃H₈
C
C₄H₁₀
D
CH₃

Answer

A. $C _2 H _6$
Explanation:
As we know,
Molecular weight $=$ Vapour density $\times 2$
$=2 \times 15=30$
Molecular formula = Empirical formula $\times n$
$n=\frac{\text { Empirical formula weight }}{\text { Molecular weight }}=\frac{30}{15}=2$
So, molecular formula $2 \times\left( CH _3\right)= C _2 H _6$

View full question & answer→

MCQ 291 Mark

The vapour density of carbon dioxide [C = 12, O = 16] is:[Board Question]

A
12
B
16
C
44
D
22

Answer

D. 22
Explanation:
The molecular formula of carbon dioxide $\left( CO _2\right)$ consists of one carbon ( C $)$ atom and two oxygen (O) atoms:
Molar mass of $C =12 g / mol$
Molar mass of $O =16 g / mol$
Molar mass of carbon dioxide $=12+2(16)=12+32=44 g$
Vapour density of gas $=\frac{\text { Molecular weight }}{2}$
$=\frac{44}{2}=22 g$
Hence, vapour density of carbon dioxide is 22 g .

View full question & answer→

MCQ 301 Mark

What is the percentage of water in $CuSO _4 \cdot 5 H _2 O$ ?

A
$12 \%$
B
$14 \%$
C
$36 \%$
D
$18 \%$

Answer

C. $36 \%$
Explanation:
Gram Molecular mass of $CuSO _4 \cdot 5 H _2 O$ is calculated as follows:
Gram molecular mass $CuSO _4 \cdot 5 H _2 O =(1 Cu +1 S+4 O )+(5(2 H +1 O )$
$\begin{array}{l}
=(1 \times 64+1 \times 32+4 \times 16) g+(5(2 \times 1+1 \times 16) g \\
=(64+32+64) g+(90) g=250 g
\end{array}$
Now, determination of mass percentage is given by the formula,
$\text { Percentage composition by mass }=\frac{\text { Mass of the element in } 1 \text { molecule of a compound }}{\text { Gram molecular mass of the compound }} \times 100$
Now
There are 5 molecules of water present as water of crystallization in one molecule of $CuSO _4 .5 H _2 O$.
The mass of one molecule of water is 18 g
The mass of water present in $CuSO _4 \cdot 5 H _2 O$ as water of crystallization is $5 \times 18 g=90 g$.
The percentage of water in crystallization in $CuSO _4 \cdot 5 H _2 O$ is calculated as follows:
$\%$ water of crystallization $=\frac{90 g}{250 g} \times 100=36 \%$
Hence, the percentage of water of crystallization in $CuSO _4 \cdot 5 H _2 O$ is $36 \%$.

View full question & answer→

MCQ 311 Mark

Naphthalene contains $93.75 \% C$ and the rest hydrogen. Molecular mass of naphthalene is 128 . Find its empirical formula.

A
$C _5 H _4$
B
$C _6 H _4$
C
$C _5 H _{10}$
D
$C _5 H _2$

Answer

A. $C _5 H _4$
Explanation:
Mass of carbon $=93.71$
Mass of hydrogen $=6.29$
Number of mole (carbon) $=\frac{93.71}{12}=7.8$
Number of mole (hydrogen) $=\frac{6.29}{1}=6.29$
Hence, most simple ratio, for carbon $=\frac{7.8}{6.29}=1.25$
For hydrogen $=\frac{6.29}{6.29}=1$
Since, it is not a whole number we multiply the ratio by 4 to get a whole number ratio.
Lowest whole number ratio is:
For carbon $=5$
For hydrogen $=4$
Empirical formula $= C _5 H _4$
Hence, the empirical formula of naphthalene is $C _2 H _4$.

View full question & answer→

MCQ 321 Mark

The empirical formula and molecular mass of a compound are CH₂O and 180 g respectively. What will be the molecular formula of the compound?

A
C₉H₁₈0₉
B
CH₂O
C
C₆H₁₂O₆
D
C₂H₄O₂

Answer

C. $C _6 H _{12} O _6$
Explanation: Given,
Empirical formula $= CH _2 O$
Empirical formula mass $=12+(1 \times 2)+16$
$=12+2+16=30 g$
Now, determining molecular formula
Given,
Molecular mass $=180 g$
Molecular formula $=\left( CH _2 O \right)_n$
Where value of $n=\frac{\text { Molecular mass }}{\text { Emprical mass }}$
$=\frac{180}{30}=6$
Thus, value of $n=6$
$\begin{array}{l}
\text { Molecular formula }=\left(CH_2 O\right)_6 \\
=C_6 H_{12} O_6
\end{array}$
Thus, for empirical formula $CH _2 O$ and molecular mass 180 g , molecular formula is $C _6 H _{12} O _6$.

View full question & answer→

MCQ 331 Mark

The correct formula showing the relation between vapour density and molecular weight:

A
Molecular weight = 2/ vapour density
B
Molecular weight $=2 \times$ vapour density
C
Molecular weight × 2 = Vapour density
D
None of these

Answer

B.Molecular weight = 2 × vapour density
Explanation: A molecular mass is the sum of the mass of all the atoms in a molecule. Whereas, vapour density is the density of vapour in relation to the density of vapour of hydrogen. The relation between molecular mass and vapour density is:
Molecular mass = 2 × vapour density

View full question & answer→

MCQ 341 Mark

The formula which gives the simple ratio of each kind of atoms present in the molecule of a compound is called:

A
Molecular Formula
B
Empirical Formula
C
Structural Formula
D
None of these

Answer

B. Empirical Formula
Explanation : Empirical Formula is defined as the formula that gives the simplest whole-number ratio of atoms in a compound.

View full question & answer→

MCQ 351 Mark

Which of the following would occupy 22.4 litres at S.T.P.? [Board Question]
1. 32g of oxygen gas
2. 2 moles of hydrogen gas
3. 6.022 x 10²³ molecules of ammonia
[Atomic weights: $O =16, H =1, N=14$ ]

A
1 & 2
B
1 & 3
C
2 & 3
D
1, 2 & 3

Answer

B. $1 \& 3$
Explanation:
Gram molecular $=1$ mole $=6.022 \times 10^{23}$ molecules $=22.4 L$
$\therefore$ Molecular Mass of $O _2$ gas $=32 g=6.022 \times 10^{23}$ molecules $=22.4 L$
Similary $6.022 \times 10^{23}$ molecule of $NH _3=22.4 L$
Since 1 mole of any gas occupies 22.4 litres at STP, 2 moles of hydrogen gas would occupy $2 \times$ $22.4=44.8$ litres, which is more than 22.4 litres.

View full question & answer→

MCQ 361 Mark

If two compounds have the same empirical formula but different molecular formulae, they must have

A
Different percentage composition.
B
Different molecular mass.
C
Same viscosity.
D
Same vapour density.

Answer

B.Different molecular mass.
Explanation: The empirical formula represents the simplest whole-number ratio of the elements in a compound, while the molecular formula specifies the actual number of atoms of each element in a molecule. If the molecular formulas are different, it means the compounds have different molecular masses.

View full question & answer→

MCQ 371 Mark

What indicates the actual number of constituent atoms in a molecule?

A
Empirical formula
B
Molecular formula
C
Empirical mass
D
Molecular mass

Answer

B. Molecular formula
Explanation: A molecular formula consists of the chemical symbols for the constituent elements followed by numeric subscripts describing the number of atoms of each element present in the molecule. Thus, the actual number of constituent atoms in a molecule is indicated by the molecular formula.

View full question & answer→

MCQ 381 Mark

The ratio of certain mass of a gas or vapour to the mass of same volume of hydrogen is it's ____________

A
Vapour density
B
Empirical formula
C
Molecular formula
D
Percentage composition

Answer

A. Vapour density
Explanation: Vapour density is the ratio of the mass of a volume of a gas, to the mass of an equal volume of hydrogen, measured under the same conditions of temperature and pressure. It is obtained by dividing the molecular weight of the vapour by the average molecular weight of air thus, it has no unit.

View full question & answer→

MCQ 391 Mark

A/An ___________ is the smallest unit of matter, which may or may not have an independent existence, but always takes part in a chemical reaction.

A
Atom
B
Molecule
C
Particle
D
Compound

Answer

A. Atom
Explanation: The smallest unit of matter that may or may not have an independent existence but always takes part in a chemical reaction is an atom.

View full question & answer→

Chemistry MCQ

Test yourself on this topic