CBSE Class 12 · 2023 · SolutionsLong Answer

WhatsApp Email

(i) Why is the boiling point of 1 M NaCl solution more than that of 1 M glucose solution?
(ii) A non-volatile solute 'X' (molar mass = 50 g mol

^{- 1}

), when dissolved in 78 g of benzene, reduced its vapour pressure to 90%. Calculate the mass of X dissolved in the solution.
(iii) Calculate the boiling point elevation for a solution prepared by adding 10 g of MgCl

_{2}

to 200 g of water, assuming MgCl

_{2}

is completely dissociated. (K

_{b}

for water = 0.512 K kg mol

^{- 1}

, molar mass of MgCl

_{2}

= 95 g mol

^{- 1}

)

Answer: (i) NaCl gives ~2 particles per unit, glucose 1 — larger ΔT_b. (ii) ≈ 5.56 g (iii) ΔT_b ≈ 0.81 K

Solution

(i) NaCl dissociates into Na

^{+}

+ Cl

^{-}

, giving nearly twice the particles of molecular glucose at the same molarity; boiling-point elevation counts particles, so the NaCl solution boils higher.
(ii) Vapour pressure reduced to 90% means

x_{1} = 0.90

, i.e.

x_{2} = 0.10

.
Moles of benzene:

n_{1} = \frac{78}{78} = 1 mol

From

x_{2} = \frac{n _{2}}{n _{1} + n _{2}}

0.10 = \frac{n _{2}}{1 + n _{2}} \Rightarrow n_{2} = \frac{1}{9} = 0.111 mol

Mass of X:

w = 0.111 \times 50

w = 5.56 g

(iii) Molality:

m = \frac{10/95}{0.200} = 0.5263 mol kg^{- 1}

Complete dissociation of MgCl

_{2}

gives

i = 3

Δ T_{b} = i K_{b} m = 3 \times 0.512 \times 0.5263

Δ T_{b} = 0.81 K

CBSE Class 12 Board · 2023 · Subjective

Practise more: all CBSE board PYQs on Solutions →

CBSE Class 12 · 2023 · SolutionsLong Answer

Solution

More board questions on Solutions