Solutions MCQs for NEET — Chemistry Questions with Answers

(b) $\frac{∆\mathrm{P}}{{\mathrm{P}}_0}={\mathrm{X}}_{\mathrm{B}}, \mathrm{so} {\mathrm{X}}_{\mathrm{B}}=\frac{760-750}{760}=\frac{1}{76}$

(d) As the no. of ionic specied produced after complete dissociation of 0.1 (M) Ca(NO₃)₂ and 0.1(M) Na₂SO₄ are same

(c) As the no. of species obtained after the complete dissociation of Al₂(SO₄)₃ is same to the no. of species obtained after complete dissociation of K₄[Fe(CN)₆]

$$\begin{gathered} \left(b\right) \mathrm{Let} \mathrm{the} \mathrm{moles} \mathrm{of} \mathrm{benzene} \mathrm{vapourizes} \mathrm{at} 20° ={\mathrm{n}}_1 \\ \mathrm{Volume} \mathrm{on} {\mathrm{n}}_1 \mathrm{mole} \mathrm{of} \mathrm{benzene} \left(\mathrm{l}\right) = \frac{78{\mathrm{n}}_1}{0.877} \\ \mathrm{Volume} \mathrm{of} {\mathrm{n}}_1 \mathrm{mole} \mathrm{of} \mathrm{benzene} \left(\mathrm{g}\right) = 2750\left(\frac{78{\mathrm{n}}_1}{0.877}\right) \\ \mathrm{PV} = \mathrm{nRT} \\ \mathrm{P}\left(\frac{78{\mathrm{n}}_1}{0.877}\right) \left(\frac{2750}{1000}\right) = {\mathrm{n}}_1 \times 0.082 \times 293 \\ {\mathrm{P}}_{\mathrm{Benzene}}^0=0.0982 \mathrm{atm} = 74.63 \mathrm{mmHg} \\ {\mathrm{P}}_{\mathrm{Benzene}}^0 \mathrm{at} 27°\mathrm{C} \mathrm{can} \mathrm{be} \mathrm{calculated} \mathrm{as} \\ \log \left(\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\right) = \frac{∆{\mathrm{H}}_{\mathrm{vap}}}{2.303 \mathrm{R}}\left[\frac{{\mathrm{T}}_2-{\mathrm{T}}_1}{{\mathrm{T}}_1{\mathrm{T}}_2}\right] \\ \log \left(\frac{{\mathrm{P}}_2}{74.63}\right) = \frac{394.57 \times 78}{2.303 \times 8.314}\left[\frac{7}{300 \times 293}\right] \\ {\mathrm{P}}_{\mathrm{Benzene}}^0 (\mathrm{at} 27°\mathrm{C}) = 100.2\mathrm{mm} \mathrm{Hg} \\ \mathrm{Molality} \mathrm{of} \mathrm{the} \mathrm{solution} = \frac{{\mathrm{X}}_{\mathrm{solute}} \times 1000}{{\mathrm{X}}_{\mathrm{solvent}} \times 78} \\ =\frac{\left({\displaystyle \frac{100.2 - 98.88}{98.88}}\right)\times 1000}{0.98 \times 78}=0.17 \\ ∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\mathrm{m} = 5 \times 0.17 = 0.85 \\ \mathrm{We} \mathrm{know} \mathrm{that} {\mathrm{K}}_{\mathrm{f}} = \left[\frac{{\mathrm{RT}}_{\mathrm{f}}^2}{1000 ∆{\mathrm{H}}_{\mathrm{f}}}\right]\mathrm{M} \\ 5=\left(\frac{8.134 \times {\mathrm{T}}_{\mathrm{f}}^2}{1000\times 10060}\right)78 \\ {\mathrm{T}}_{\mathrm{f}} =278.5\mathrm{K} \\ \therefore \mathrm{Freezing} \mathrm{point} \mathrm{of} \mathrm{solution} = 278.5 - 0.85= 277.65 \mathrm{K} \end{gathered}$$

(d)

Mass of the acid = 10 g

Mass of water = 250 g

Mass pf acid per kg of water = $\frac{10 \mathrm{g} \times 1000\mathrm{g}}{250 \mathrm{g}}=40 \mathrm{g}$

Molar mass of the acid  = 122.5 g/mol

The given acid is a weak acid. So, 

$$\begin{gathered} \mathrm{\alpha }=\sqrt{\frac{{\mathrm{K}}_{\mathrm{a}}}{\mathrm{C}}}=\sqrt{\frac{1.4 \times {10}^{-3}}{\left({\displaystyle \raisebox{1ex}{$40$}\!\left/ \!\raisebox{-1ex}{$122.5$}\right.}\right)}}=0.065 \\ \mathrm{So}, \mathrm{i}=\mathrm{\alpha }(\mathrm{n}-1)+1=0.065(2-1)+1=1.065 \\ \mathrm{Then}, ∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{iK}}_{\mathrm{f}}\mathrm{m}=1.065 \times 1.86\times \frac{40}{122.5}=0.65 \mathrm{K} \\ \mathrm{SO}, \mathrm{the} \mathrm{depression} \mathrm{in} \mathrm{the} \mathrm{freezing} \mathrm{of} \mathrm{water} \mathrm{will} \mathrm{be} 0.65°\mathrm{C}. \end{gathered}$$

(a) This is one of forms of statements of Raoult's law

$\frac{\mathrm{P}-{\mathrm{P}}_{\mathrm{s}}}{\mathrm{P}}=\frac{\mathrm{n}}{\mathrm{n}+\mathrm{N}}$

Where n=number of moles of solute, P and P_s are the vapour pressure of the solvent and solution.

(a) 

$$\begin{gathered} \mathrm{P}={\mathrm{P}}_{\mathrm{A}}+{\mathrm{P}}_{\mathrm{B}} \\ {\mathrm{P}}_{\mathrm{A}}={\mathrm{X}}_1 \mathrm{P}={\mathrm{X}}_2{\mathrm{P}}_{\mathrm{A}}^0 \\ \therefore \mathrm{P}=\frac{{\mathrm{P}}_{\mathrm{A}}^0{\mathrm{X}}_2}{{\mathrm{X}}_1} \end{gathered}$$

$$\begin{gathered} \mathrm{HA}\stackrel{ }{\rightleftharpoons }{\mathrm{H}}^{+}+{\mathrm{A}}^{-} \mathrm{C}(1+\mathrm{\alpha }) \\ \mathrm{C}(1-\mathrm{\alpha }) \mathrm{C\alpha } \mathrm{C\alpha } \\ \left[{\mathrm{H}}^{+}\right]={10}^{-2} \left[{\mathrm{A}}^{-}\right]={10}^{-2} \\ \left[\mathrm{HA}\right]=0.1-{10}^{-2}=0.09 \\ \mathrm{C}(1-\mathrm{\alpha })=0.09 \\ (1-\mathrm{\alpha })=\frac{0.09}{0.1}=0.9 \\ \mathrm{\alpha }=1-0.9=0.1 \\ \therefore \mathrm{\pi }=\mathrm{C}(1+\mathrm{\alpha })\mathrm{RT}=0.1(1.1)\mathrm{RT}=0.11\mathrm{RT} \end{gathered}$$

(c)$\frac{23.76-23.392}{23.76}=\frac{\mathrm{w}}{342\times 5.65}$

w= 30 g

(a) This depends on the number of ionic species present.