Chemistry M.C.Q (1 Marks)

$ \text { Slope }=R T $

$ 25.73=0.083 \times T $

$T=\frac{25.73}{0.083}=309.47 \approx 310 \mathrm{~K}$

$\therefore \quad \text { Temperature in }{ }^{\circ} \mathrm{C} $$ =310-273 $

$ =37^{\circ} \mathrm{C}$

Higher the value of $\mathrm{K}_{\mathrm{H}}$ at a given pressure, lower is the solubility of the gas in the liquid.

$\mathrm{K}_H$ value of gases (given) : $A>C>B$

$\therefore \quad$ Order of solubility of gases in water : $\mathrm{B}>\mathrm{C}>\mathrm{A}$

$\Pi \propto \mathrm{C} \Rightarrow \Pi \propto \frac{1}{\mathrm{M}_{\mathrm{w}}}$

Glucose $\Rightarrow \mathrm{M}_{\mathrm{w} 1}=180$

Urea $\Rightarrow \mathrm{M}_{\mathrm{w} 2}=60$

Sucrose $\Rightarrow \mathrm{M}_{\mathrm{w} 3}=342$

$\mathrm{P}_{2}>\mathrm{P}_{1}>\mathrm{P}_{3}$

$=280 \times \frac{3}{5}+420 \times \frac{2}{5}$

$=56 \times 3+84 \times 2$

$=168+168$

$=336$

$=5.12 \times 0.078$

$\Delta T _{ f }=0.40 K$

$\frac{\Delta P}{P_{A}^{0}}=\frac{n_{B}}{n_{A}}=\frac{w_{B}}{m_{B}} \cdot \frac{m_{A}}{W_{A}}$

$\frac{20}{100}=\frac{8}{ m _{ B }} \cdot \frac{114}{114}$

$m _{ B }=\frac{8 \times 100}{20}=40 gmol ^{-1}$

$0.835=1.67 \times 10^{3} \times \frac{\mathrm{n}\left(\mathrm{CO}_{2}\right)}{\frac{0.9 \times 1000}{18}}$

$\mathrm{n}\left(\mathrm{CO}_{2}\right)=0.025$

Millimoles of $\mathrm{CO}_{2}=0.025 \times 1000=25$

$=37 \times 0.5+119 \times 0.5$

$=78$

$P _{ A }= Y _{ A } P _{\text {total }}$

$\therefore y _{ A }=\frac{ P _{ A }}{ P _{\text {total }}}=\frac{ P _{ A }^{0} X _{ A }}{ P _{\text {total }}}=\frac{37 \times 0.5}{78}=0.237$

Since the enthalpy of mixing (solution) is zero, the change in Gibbs energy on mixing is determined solely by the entropy of mixing ( $\Delta S _{\text {solution }}$ ).

So the $\Delta G$ is not zero.

$i=3$

$p_{s}=732\, \mathrm{mm},\,\, K_{b}=0.52$

$T_{b}^{o}=100\,^{o} \mathrm{C},\,\, p^{o}=760 \,\mathrm{mm}$

$\frac{p^{o}-p_{s}}{p^{o}}=\frac{n_{2}}{n_{1}}$

$\Rightarrow \frac{760-732}{760}=\frac{n_{2}}{100 / 18}$

$\Rightarrow n_{2}=\frac{28 \times 100}{760 \times 18}=0.2046\, \mathrm{mol}$

$\Delta T_{b}=K_{b} \times m$

$T_{b}-T_{b}^{o}=K_{b} \times \frac{n_{2} \times 1000}{W_{A}\,(g)}$

$ T_{b}-100\,^{o} \mathrm{C} =\frac{0.52 \times 0.2046 \times 1000}{100} $

$=1.06 $

$T_{b}=100+1.06 =101.06\,^{o} \mathrm{C} $

$\because \mathrm{YA}=\frac{\mathrm{P}^{o} \mathrm{A} \mathrm{X} \mathrm{A}}{\mathrm{P}}, \mathrm{YB}=\frac{\mathrm{P}^{o} \mathrm{BXB}}{\mathrm{P}}$

$\mathrm{X}_{\mathrm{A}}$ and $\mathrm{X}_{\mathrm{B}}$ are same so mole fraction in vapour phase is directly proportional to the vapour pressure.

$\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3} \rightarrow \mathrm{Al}^{3+}+3 \mathrm{NO}_{3}^-, i=4$

$\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \rightarrow 4 \mathrm{K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{-4}, i=5$

$\mathrm{K}_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{K}^{+}+\mathrm{SO}_{4}^{2-}, i=3$

$\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \rightarrow 3 \mathrm{K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}, i=4$

For equimolal solutions, elevation in boiling point will be higher if solution undergoes dissociation i.e., $i>1$

(i) for $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \longrightarrow \mathrm{i}=5$

$\Delta_{\operatorname{mixing}} \mathrm{H}=0 ; \Delta_{\operatorname{mixing}} \mathrm{V}=0$

and it should obey Raoult's law.

$=0.5$ moles No. of moles $=$ molarity $\times$ volume

Requird mass of $\mathrm{HNO}_{3}=0.5 \times 63$

$=31.5 \mathrm{gm}$

$70 \mathrm{gm}$ of $\mathrm{HNO}_{3}$ are present in $100 \mathrm{gm}$ of solution,

So $1\; gm$ will be present in $100 / 70 \mathrm{gm}$ of solution.

$31.5 \mathrm{gm}$ be present in $(100 / 70) \times 31.5 \mathrm{gm}$ of solution

$=45 \;\mathrm{gm}$

Initial conc. $1\, m\quad \quad \quad \quad \quad \quad 0\quad \quad \quad \quad 0$

Final conc. $(1-0.4) \,m \quad \quad 4 \times 0.4 \quad \quad 0.4\, m$

$\quad \quad \quad \quad \quad \quad =0.6\,m \quad \quad \quad =1.6\,m$

Effective molality $=0.6+1.6+0.4=2.6 \,m$

For same boiling point, the molality of another solution should also be $2.6\, m$.

Now,$18.1$ weight percent solution means $18.1\, gm$ solute is present in $100\, gm$ solution and hence, $(100-18.1=) 81.9 \,gm$ water.

$\text { Now, } \quad 2.6=\frac{18.1 / M }{81.9 / 1000}$

$\therefore$ Molar mass of solute, $M =85$

$\alpha=0.05=50 \times 10^{-3}$

$0.1 \,M\, NaOH$

$\left( OH ^{-}\right)=0.1+0.05\,\, \alpha \,\approx \,0.1$

$K _{ b }=\frac{[ C \alpha]\left[ OH ^{-}\right]}{[ C - C \alpha]}$

$5 \times 10^{-4}=\frac{\alpha \times 0.1}{1-\alpha}$

$\alpha=5 \times 10^{-1} \,\%$

Before association $1$ mole

After association $1-x \quad x / 2$

Total $=1-x+\frac{x}{2}=1-\frac{x}{2}$

$\therefore i=\frac{1-x / 2}{1}$

or, $i=1-\frac{x}{2}$

$0.003=3 \times 5 \times S$

$\mathrm{S}=2 \times 10^{-4}$

$\mathrm{K}_{\mathrm{sp}}=4 \mathrm{S}^{3}=3.2 \times 10^{-11}$

$0.75=i \times 5 \times 0.1$

$i=1.5$   ;          $\alpha=\frac{1.5-1}{2-1}=0.5$

$\mathrm{K}_{\mathrm{a}}=\frac{\alpha^{2} \mathrm{c}}{1-\alpha}=\frac{(0.5)^{2} \times 0.1}{1-0.5}=5 \times 10^{-2}$

$\mathrm{i} \times \mathrm{m}$ of $\mathrm{Na}_{2} \mathrm{SO}_{4}$ is highest,

hence its boiling point will also be highest.

$\mathrm{H}^{+}=10^{-2}=\mathrm{C} \alpha(\text { mono-basic acid })$

${\alpha=\frac{10^{-2}}{0.1}=0.1} $

${\mathrm{i}=1+\alpha(\mathrm{N}-1)}$

${\mathrm{i}=1+0.1(2-1)}$

${\mathrm{i}=1.1}$

${\pi=0.1 \times \mathrm{RT} \times 1.1} $

${\pi=0.11\,\, \mathrm{RT}}$

$1 \quad\quad\quad 0$

$1-\infty \quad \frac{\alpha}{n}$

Total moles $=1-\alpha+\frac{\alpha}{n}$

$i=\frac{\text { Total moles at eq }}{\text { Intial moles }}=\frac{1-\alpha+\frac{\alpha}{n}}{1}$

$5 \,\mathrm{W}=\mathrm{W}+60$

$4\, \mathrm{W}=60$

$\mathrm{W}=15\, \mathrm{g}$

               $=2 \times 0.51 \times 0.1$

$\Delta \mathrm{T}=0.102$

$\mathrm{T}_{\mathrm{b}}=100+0.102=100.1\,^{o} \mathrm{C}$

$0.9=\frac{i-1}{2-1} $       $\therefore \quad i=1.9 $

${\pi =i C R T=1.9 \times 0.002 \times 0.082 \times 300}$

${=0.094\, \text { bar }}$

$1 - \alpha $    $\frac{\alpha }{2}$

$1 - 0.8$    $\frac{{0.8}}{2}$

$i\, = 1 - 0.8\, + \,\frac{{0.8}}{2}\, = \,0.6$

$\Delta {T_f}\, = \,{K_f}\, \times \,\,i\,\, \times \,\,m$ $ = \,5\,\, \times \,0.6\,\, \times \,\frac{X}{{122}}\, \times \,\frac{{1000}}{{30}}\, = \,2$ (Since $\Delta {T_f}\, = \,2$ )

$\therefore \,X\, = \,2.44\,g$

Total number of particle $=\,1+2\alpha $

Hence, Van't Hoff factor $=\,\frac {1+2\alpha }{1}$

$=\frac {1+2\times 0.4}{1}$ $=\,1+0.8$ $\Rightarrow $ $1.8$

$ = \,\frac{{\Delta {T_f}}}{{{K_f}}}\, = \,\frac{{3.82}}{{1.86}}\, = \,2.054\,mol/1000\,g$ solvent

Molality (theoretical)

$ = \,\frac{{mole\,\,of\,solute}}{{wt.\,of\,solventing\,(g)}} \times 1000$

$ = \,\frac{{5\,g\,/\,142\,g\,/\,mole}}{x} \times 1000$

                                            $N{a_2}S{O_4}\, \to \,2N{a^ + }\, + \,SO_4^{2 - }$

Moles before dissociation              $1$              $0$              $0$

Moles after dissociation              $1-x$              $2x$              $x$

Von't Hoff Factor $(i) = \,\frac{{Moles\,after\,\,dissociation}}{{Moles\,before\,\,dissociation}}$

                               $ = \frac{{(1 - x)\, + \,2x\, + \,x}}{1}$

$Na_2SO_4$ is ionised $81.5\%$ means $x\,=\,0.815$

                             $ = \frac{{(1 - 0.815)\, + \,2 \times 0.815\, + \,0.815}}{1}$

                              $=\,2.63$

$i\, = \,\frac{{Observed\,\,molarrity}}{{Calculated\,\,molarity}}$

$ \Rightarrow \,2.63\, = \,\frac{{2.054}}{{\frac{{0.0352}}{x} \times 1000}}\, = \,45.07\,g$

$\because \pi_{1}=\frac{w}{m . V} S T(1+2 \alpha)$

$\pi_{1}=\frac{17.4 \times 1000}{174 \times 100} \times S T(1+2 \alpha)$

$=S T \times(1+2 \alpha)$

$\begin{array}{cll} NaCl \rightleftharpoons & Na^{+}+Cl^- \\ 1 & 0 \;\;\;\;\;\;\;\;\;\;\;\; 0 \\ 1-\alpha_1 & \alpha_ 1 \;\;\;\;\;\;\;\;\; \alpha_1\end{array}$

$\therefore \pi_{2}=\frac{5.85 \times 1000}{58.5 \times 100} \times S T \times\left(1+\alpha_{1}\right)$

$\because \alpha_{1}=1$

$\therefore \pi_{2}=S T \times 2$

Given $\pi_{1}=\pi_{2}$ (isotonic solution) $\therefore S T \times 2=S T \times(1+2 \alpha)$

or $\alpha=0.5$

or $50 \%$ ionisation of $K_{2} S O_{4}$

$\Rightarrow \frac{\pi( KI \text { solution })}{\pi \text { (sucrose solution) }}=\frac{0.432 atm }{0.24 atm }=1.80$

$\Rightarrow \alpha=0.889=88.9 \%$

$\text { For } \mathrm{NaOH} \Rightarrow \alpha=100 \% \quad \mathrm{i}=2$

$\mathrm{i} \times \frac{17.4 \times 1000}{174 \times 100}$ $=\frac{2 \times 4 \times 1000}{40 \times 100}$

$i=2$

$\text { For } \mathrm{K}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{K}^{+}+\mathrm{SO}_{4}^{2-} \Rightarrow \mathrm{n}=3$

$\mathrm{i}=1+\alpha(\mathrm{n}-1)$

$2=1+\alpha(3-1)$

$2=1+2 \alpha$

$1=2 \alpha$

$\alpha=1 / 2=$

${\alpha=0.5}$

${\alpha=50 \%}$

$=1+(2-1) 0.2$

$=1.2$

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{K}_{\mathrm{f}} \mathrm{m}$

$=1.2 \times 1.86 \times 0.5$

$=1.12\, \mathrm{K}$

$\mathrm{i}=1+\left(\frac{1}{3}-1\right) \times 0.8=0.47$

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{K}_{\mathrm{f}}$ molality

${=0.47 \times 5.12 \times 0.25} $

$=\,0.6{{\,}^{o}}C$

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{T}_{\mathrm{o}}-\mathrm{T}_{\mathrm{s}}$

$0.6=5.5-\mathrm{T}_{\mathrm{s}}$

$\mathrm{T}_{\mathrm{s}}=4.9\,^{o} \mathrm{C}$ or $278\, \mathrm{K}$

Moles of $\left[\mathrm{Na}^{+}\right]$ ions $=0.2 \times 0.5=0.1$

Concentration of $\left[\mathrm{Na}^{+}\right]$ ions $=\frac{0.1}{2}=0.05 \,\mathrm{M}$

Moles of $\left[\mathrm{Mg}^{2+}\right]$ ions $=0.4 \times 1.5=0.6$

Concentration of $\left[\mathrm{Mg}^{2+}\right]$ ions

$=\frac{0.6}{2}=0.3\, \mathrm{M}=7.2\, \mathrm{gm} / \mathrm{L}$

Moles of $\left[\mathrm{Cl}^{-}\right]$ ions $=0.2 \times 0.5+2 \times 0.4 \times 1.5$

                                $=1.3$

Concentration of $[\mathrm{Cl}^-]$ ions $=\frac{1.3}{2}=0.65\, \mathrm{M}$

$n=3$

$\alpha=\frac{i-1}{n-1}$

$0.6=\frac{i-1}{3-1}$

$i=2.2$

${\mathrm{n}=3} $

${\mathrm{i}=1+(\mathrm{n}-1) \alpha}$

${\mathrm{i}=1+(3-1) \times 0.8} $

${0.8=\frac{\mathrm{i}-1}{3-1}}$

${\mathrm{i}=2.6}$