An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Vapour pressure of the solution at normal boiling point (p1) = 1.004 bar
Vapour pressure of pure water at normal boiling point (`p_1^0=1.013 " bar"`)
Mass of solute, (w2) = 2 g
Mass of solvent (water), (w1) = 98 g
Molar mass of solvent (water), (M1) = 18 g mol−1
According to Raoult’s law,
`(p_1^0-p_1)/p_1^0 = (w_2xxM_1)/(M_2xxw_1)`
`=>0.009/1.013 = (2xx18)/(M_2xx98)`
`=>M_2 = (1.013xx2xx18)/(0.009xx98)`
= 41.35 g mol-1
Hence, the molar mass of the solute is 41.35 g mol−1.