Solutions MCQs for NEET — Chemistry Questions with Answers

(a) Given,

molality, m=0.0020 m

             $$\begin{gathered} ∆{\mathrm{T}}_{\mathrm{f}}=0^{\circ }\mathrm{C}-(- 0.00732°\mathrm{C}) \\ = 0 + 0.00732°\mathrm{C} \\ = 0.00732°\mathrm{C} \\ {\mathrm{k}}_{\mathrm{f}} = 1.86°\mathrm{C}/\mathrm{m} \\ ∆{\mathrm{T}}_{\mathrm{f}}=\mathrm{i}·{\mathrm{k}}_{\mathrm{f}}\times \mathrm{m} \\ \mathrm{i}=\frac{∆{\mathrm{T}}_{\mathrm{f}}}{{\mathrm{k}}_{\mathrm{f}}\times \mathrm{m}}= \frac{0.00732}{1.86 \times 0.0020}= 1.96 \simeq 2 \end{gathered}$$

Since, the compound is ionic, so number of moles produced is equal to vant' Hoff factor, i. Hence, 2 moles of ions are produced.

$$\begin{gathered} \left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5{\mathrm{NO}}_2\right]\mathrm{Cl} \stackrel{ }{\to }{\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5{\mathrm{NO}}_2\right]}^{+} + {\mathrm{Cl}}^{-} \\ 1 \mathrm{mol} 2 \mathrm{ions} \end{gathered}$$

(d) According to Kohlrausch's law "at infinite dilution when the dissociation is complete, each ion makes a definite contribution towards equivalent conductivity of the electrolyte irrespective of the nature of the other ion with which it is associated

    Or

Equivalent conductivity of an electrolyte at infinite dilution is the sum of the equivalent conductivities of the cations and anion.

HX $\rightleftharpoons$H⁺ + X^-

1            0           0

1-$\mathrm{\alpha }$         $\mathrm{\alpha }$           $\mathrm{\alpha }$      (at equilibrium)

$\mathrm{\alpha }$ = 20% dissociation

i.e., $\mathrm{\alpha }$ = 0.2

i = 1-$\mathrm{\alpha }$+$\mathrm{\alpha }$+$\mathrm{\alpha }$

=1+$\mathrm{\alpha }$=1+0.2=1.2

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

= 1.2 x 1.86K kg mol^-1 x 0.5

= 1.12 K

10 g per dm³ of urea is isotonic with 5% solution of a non-volatile solute. Hence, between these solution osmosis is not possible so their molar concentrations are equal to each other,

Thus, molar concentration of urea solution = (10 g/dm³)/Mol.wt of urea = 10/60 M = 1/6 M

Molar concentration of 5% non-volatile solute = (50 g/dm³)/mol wt of non-volatile solute = 50/m M

Both solutions are isotonic to each other, therefore

 1/6 = 50/m

or m = 50 x 6 = 300 g mol^-1

(b) A solution of acetone in ethanol shows a positive deviation from Raoult's law due to miscibility of these two liquids with difference of polarity and length of hydrocarbon chain.

During osmosis, flow of water through a semi-permeable membrane is from solution having lower concentration only. 

The loss in weight should be proportional to vapour pressure above that solution:

So, ${\text{P}}_{{\text{S}}_{\text{A}}}\propto$ 2gm

${\text{P}}_{{\text{S}}_{\text{B}}}\propto$ 1.5gm

${\text{P}}_{{\text{S}}_{\text{C}}}\propto$ 2.5gm

So, maximum vapour pressure is above C solution hence, it is having minimum lower and hence minimum mole fraction (hence minimum number of moles of solute) So max, molar mass of substance.

The molar heat of vaporization (ΔHv) can be calculated using the Clausius-Clapeyron equation: ln(P2/P1) = -(ΔHv/R) * ((1/T2) - (1/T1)). Substituting the given values and using log(5) = 0.3 * log(2), we get: ΔHv = (8.314 * 0.3 * (356 - 315)) / (1/356 - 1/315) = 37.5 kJ/mol.

$\frac{\Delta \text{P}}{\text{P}}{\text{\hspace{0.17em}X}}_{\text{solute}}=\text{\hspace{0.17em}i\hspace{0.17em}}\times \frac{2}{5}=\frac{1}{2}$

so, i = 1.25

⇒ Hence degree of dissociation = $\frac{1}{4}$

so,moles of Cl^– ions 2 × $\frac{1}{4}$= $\frac{1}{2}$moles

so, moles of AgCl ppt = $\frac{1}{2}$moles = 0.5 mol AgCl

Higher freezing point ⇒ lesser ΔT_f

⇒ lesser molality

⇒ lesser number of particles