18 glucose (C6 H12 O6) is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution at 273 K is

Question

18 glucose (C6 H12 O6) is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution at 273 K is (A) 759.0g (B) 739.6g (C) 746.0g (D) 752.4g

Tardigrade Admin · Accepted Answer

Relative lowering of vapour pressure Equation (P0 - Ps/P0)=Xs o l u t e=(n/n + N) Modified forms of equation is (P0 - Ps/Ps)=(n/N) n = moles of solute, N = moles of solvent P0=760 torr Ps=? (760 - Ps/Ps)=((18/180)/(178.2)18) Ps=752.4 torr

Q. 18 glucose (C6​H12​O6​) is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution at 273 K is

759.0g

739.6g

746.0g

752.4g

Relative lowering of vapour pressure Equation P0P0−Ps​​=Xsolute​=n+Nn​ Modified forms of equation is Ps​P0−Ps​​=Nn​ n = moles of solute, N = moles of solvent P0=760 torr Ps​=? Ps​760−Ps​​=18178.2​18018​​ Ps​=752.4 torr

Q. 18 glucose $(C_{6} H_{12} O_{6})$ is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution at 273 K is