Determine the vapor pressure of a solution at 25â°c that contains 76.6 g of glucose (c6h12o6) in 250.0 ml of water. the vapor pressure of pure water at 25â°c is 23.8 torr.70.8 torr7.29 torr72.9 torr22.9 torr23.1 torr

The vapor pressure of solution is 23.1 Torr

Explanation:

Colligative property of vapour pressure.

ΔP = P° . Xm

ΔP = Pressure of vapour from pure solvent - Pressure of vapour from solution

ΔP = P° - Ps

Xm = mole fraction of solute

Molar weight of glucose: 180 g/m

Moles of glucose = mass / molar weight → 76.6 g / 180 g/m = 0.425moles

If water volume is 250 mL, this volume is occupied by 250 g of solvent.

Density of water : 1g/mL

Molar weight of water: 18 g/m

250 g / 18g/m = 13.89 moles of solvent

Total moles = moles of solute + moles of solvent

0.425 + 13.89 = 14.315 moles

Mole fraction of solute = 0.425 / 14.315 = 0.029

23.8 Torr - Ps = 23.88 Torr . 0.029

Ps = 23.88 Torr . 0.029 - 23.8 Torr ⇒ 23.1 Torr

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.