The vapour pressure of pure water at 26° C is 25.21 torr. What is the vapour pressure of a solution which contains 20.0 g glucose, C6H12O6 in 70 g water?

Question

The vapour pressure of pure water at 26° C is 25.21 torr. What is the vapour pressure of a solution which contains 20.0 g glucose, C6H12O6 in 70 g water? (A) 2.79 torr (B) 15.7 torr (C) 24.5 torr (D) 32.0 torr

Tardigrade Admin · Accepted Answer

Moles of glucose =(20.0 g glu cos e/180 g text/mol)=0.111 mol Moles of water =(70.0 g H2O/18.0g/mol)=3.89 mol H2O x(H2O)=(3.89/0.11+3.89)=0.972 p=p° xH2O=(25.21 texttorr) (0.972)=24.5 texttorr

Q. The vapour pressure of pure water at $26^{\circ} C$ is 25.21 torr. What is the vapour pressure of a solution which contains 20.0 g glucose, $C_{6} H_{12} O_{6}$ in 70 g water?

2.79 torr

15.7 torr

24.5 torr

32.0 torr

Moles of glucose $= \frac{20.0 g g l u c o s e}{180 g / m o l} = 0.111 m o l$ Moles of water $= \frac{70.0 g H _{2} O}{18.0 g / m o l} = 3.89 m o l H_{2} O$ $x (H_{2} O) = \frac{3.89}{0.11 + 3.89} = 0.972$ $p = p^{\circ} x_{H_{2} O} = (25.21 torr) (0.972) = 24.5 torr$

Q. The vapour pressure of pure water at 26∘C is 25.21 torr. What is the vapour pressure of a solution which contains 20.0 g glucose, C6​H12​O6​ in 70 g water?