Glucose of 18 g was dissolved in 170 g of water. The vapour pressure of water for this aqueous solut

Question

Glucose of 18 g was dissolved in 170 g of water. The vapour pressure of water for this aqueous solution at $37^{\circ} \mathrm{C}$ is [ Given : vapour pressure of water as solvent is 760 torr]

Professor PC Thomas Classes · Answer

Moles glucose $= 18/180 = 0.1$. Moles water $= 170/18 \approx 9.44$. $P = P^0 	imes X_{	ext{solvent}} = 760 	imes \frac{9.44}{9.54} \approx 752$ torr.