36 g glucose ( C 6 H 12 O 6) is added to 176.4 g water. The vapour pressure of water (in torr) for this aqueous solution at 100° C is.

Question

36 g glucose ( C 6 H 12 O 6) is added to 176.4 g water. The vapour pressure of water (in torr) for this aqueous solution at 100° C is. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Molecular weight of glucose ( C 6 H 12 O 6)=180 ∴ Number of moles of glucose (nB)=(36/180)=0.2 Molecular weight of water ( H 2 O )=18 ∴ Number of moles of water ( n A )=(176.4/18)=9.8 ∴ Mole fraction of water (x H 2 O )=( n A / n A + n B ) =(9.8/9.8+0.2) =(9.8/10) ∵ p H 2 O =x H 2 O p H 2 O ° ∴ p H 2 O =(9.8/10) × 760=744.8 torr (At boiling point of water, it vapour pressure is 1 atm =760 mm Hg )

Q. 36g glucose (C6​H12​O6​) is added to 176.4g water. The vapour pressure of water (in torr) for this aqueous solution at 100∘C is_______.

Q. $36 g$ glucose $(C_{6} H_{12} O_{6})$ is added to $176.4 g$ water. The vapour pressure of water (in torr) for this aqueous solution at $10 0^{\circ} C$ is_______.