6. ⇒ (MHT CET 2021 21th September Morning Shift )

What is boiling point of a decimolal aqueous solution of glucose if molal elevation constant for water is 0.52 $^{\circ}$ C kg mol $^{- 1}$ ?

A. 101.52 $^{\circ}$ C

B. 99.95 $^{\circ}$ C

C. 99.48 $^{\circ}$ C

D. 100.052 $^{\circ}$ C

Correct Option is (D)

$\begin{aligned} m = 0.1 m, K_{b} = {0.52}^{\circ} C kg {mol}^{- 1} \\ Δ T_{b} = K_{b} \times m = {0.52}^{\circ} C kg {mol}^{- 1} \times 0.1 mol {kg}^{- 1} = 0.052^{\circ} C \\ Δ T_{b} = T_{b} - T_{b}^{0} ∴ T_{b} = Δ T_{b} + T_{b}^{0} \\ ∴ T_{b} = 0.052 + 100 = 100.052^{\circ} C \end{aligned}$

7. ⇒ (MHT CET 2021 20th September Evening Shift )

Which of the following statements is correct for boiling point of a liquid?

A. Temperature at which a liquid boils at any pressure.

B. Temperature at which solid is in equilibrium with its liquid.

C. Temperature at which vapour pressure equals the applied pressure.

D. Temperature at which applied pressure is greater than vapour pressure of liquid.

Correct Option is (C)

Currently no explanation available

8. ⇒ (MHT CET 2021 20th September Morning Shift )

A solution of 6 g of solute in 100 g of water boils at 100.52 $^{\circ}$ C. The molal elevation constant of water is 0.52 K kg mol $^{- 1}$ . What is molar mass of solute?

A. 60 g mol $^{- 1}$

B. 120 g mol $^{- 1}$

C. 90 g mol $^{- 1}$

D. 180 g mol $^{- 1}$

Correct Option is (A)

W $_{2}$ = 6 g, W $_{1}$ = 100 g, K $_{b}$ = 0.52 K kg mol $^{- 1}$ , T $_{b}$ = 100.52 $^{\circ}$ C

$∴$ $Δ T_{b} = T_{b} - T_{b}^{o} = (100.52 + 273) - (100 + 273) = 0.52 K$

Now,

$\begin{array}{r} M_{2} = \frac{1000 \cdot K_{b} \cdot W_{2}}{Δ T_{b} \cdot W_{1}} = \frac{1000 g {kg}^{- 1} \times 0.52 K kg {mol}^{- 1} \times 6 g}{0.52 K \times 100 g} \end{array}$

$∴ M_{2} = 60 g {mol}^{- 1}$

⇒Solution