18 g of glucose $\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)$ is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at $100^{\circ} \mathrm{C}$ is

JEE Main · 2006 · 30 Apr 2006 · Offline · Solutions

18 g of glucose

(C_{6} H_{12} O_{6})

is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at

10 0^{\circ} C

(A)759.00 Torr
(B)7.60 Torr
(C)76.00 Torr
(D)752.40 Torr✓ Correct

Answer: (D)

Solution

Relative lowering of vapour pressure follows

\frac{P ^{\circ} - P _{s}}{P _{s}} = \frac{n}{N}

in terms of the solute and solvent moles. Glucose supplies

\frac{18}{180} = 0.1

mol and water

\frac{178.2}{18} = 9.9

mol, while pure water at

10 0^{\circ} C

has

P^{\circ} = 760 Torr

. Then

\frac{760 - P _{s}}{P _{s}} = \frac{0.1}{9.9}

, so

760 = P_{s} (1 + \frac{1}{99}) = \frac{100}{99} P_{s}

, giving

P_{s} = \frac{760 \times 99}{100} = 752.4 Torr

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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