The vapour pressure of water at 20° is 17.5 mm Hg If 18 g of glucose (C6H12O6) is added to 178.2 g of water at 20°C, the vapour pressure of the resulting solution will be:

Question

The vapour pressure of water at 20° is 17.5 mm Hg If 18 g of glucose (C6H12O6) is added to 178.2 g of water at 20°C, the vapour pressure of the resulting solution will be: (A) 17.675 mmHg (B) 15.750 mmHg (C) 16.500 mmHg (D) 17.325 mmHg

Tardigrade Admin · Accepted Answer

(P°-Ps/Ps)=(w × M/m × W) (17.5-PS/PS)=(18×18/180×178.2) (17.5-PS/PS)=0.01 PS=17.325 mmHg

Q. The vapour pressure of water at $2 0^{\circ}$ is $17.5 mm H g$ If $18 g$ of glucose $(C_{6} H_{12} O_{6})$ is added to $178.2 g$ of water at $2 0^{\circ} C$ , the vapour pressure of the resulting solution will be:

17.675 mmHg

15.750 mmHg

16.500 mmHg

17.325 mmHg

$\frac{P ^{\circ} - P _{s}}{P _{s}} = \frac{w \times M}{m \times W}$
$\frac{17.5 - P _{S}}{P _{S}} = \frac{18 \times 18}{180 \times 178.2}$
$\frac{17.5 - P _{S}}{P _{S}} = 0.01$
$P_{S} = 17.325 mm H g$

Q. The vapour pressure of water at 20∘ is 17.5mmHg If 18g of glucose (C6​H12​O6​) is added to 178.2g of water at 20∘C, the vapour pressure of the resulting solution will be: