Chemistry MCQs for NEET — Practice Questions with Answers

Key Idea: Vapour pressure depends upon the surface area of the solution. Larger the surface area, higher is the vapour pressure.

Addition of solute decreases the vapour pressure as some sites of the surface are occupied by solute particles, resulting in decreased surface area. However, addition of solvent, i.e., dilution, increases the surface area of the liquid surface, thus results in increased vapour pressure.

Hence, addition of water to the aqueous solution of (1 molal) KI, results in increased vapour pressure.

Depression in freezing point

$∆$T_f = k_f x m

where, m = molality = w_B x 1000/M_B.W_A = 68.5 x 1000/342 x 1000 = 68.5/342

 $∆$T_f = 1.86 x 68.5/342 = 0.372$°$C

$∆$T_f = T$°$-T_s = 0-0.372$°$C

Key Idea ${\mathrm{\lambda }}_{\mathrm{eq}} = \mathrm{k} \mathrm{x} \mathrm{v} = \left(\mathrm{kx}1000\right)\mathrm{normality}$

On dilution, the number of current carrying particles per cm³ decreases but the volume of solution increases. Consequently, the ionic mobility increases, which in turn increases the equivalent conductance of strong electrolyte.

Degree of dissociation, α=^^c/^∞ where, ^^c and ^∞ are equivalent conductances at a given concentration and at infinite dilution respectively.
    ⇒ α=8.0/400=2x10^-2

From Ostawald’s dilution law (for weak monobasic acid)

K_a = Cα/(1- α) = Cα² (∴ 1>>>α) = 1/32 (2x10^-2)² = 1.25x10^-5

(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.