Lowering of vapour pressure of an aqueous solution of a non-volatile, non-electrolyte 1 molal aqueous solution at 100° C is:

Question

Lowering of vapour pressure of an aqueous solution of a non-volatile, non-electrolyte 1 molal aqueous solution at 100° C is: (A) 14.12 torr (B) 312 torr (C) 13.45 torr (D) 352 torr

Tardigrade Admin · Accepted Answer

P 0=760 torr; 1 molal aqueous solution contains 1 mole of solute in 1000 g (55.5 mol ) of water. xB=(1/1+55.5)=(1/56.5) (Δ P/P)=XB, Δ P=p XB ⇒ (760 × 1/56.5)=13.45 torr

Q. Lowering of vapour pressure of an aqueous solution of a non-volatile, non-electrolyte 1 molal aqueous solution at $10 0^{\circ} C$ is:

14.12 torr

312 torr

13.45 torr

352 torr

$P^{0} = 760$ torr; 1 molal aqueous solution contains 1 mole of solute in $1000 g (55.5 m o l)$ of water.

$x_{B} = \frac{1}{1 + 55.5} = \frac{1}{56.5}$

$\frac{Δ P}{P} = X_{B}, Δ P = p X_{B}$

$\Rightarrow \frac{760 \times 1}{56.5} = 13.45$ torr

Q. Lowering of vapour pressure of an aqueous solution of a non-volatile, non-electrolyte 1 molal aqueous solution at 100∘C is: