A solution of sucrose is prepared by dissolving 68.4 g in 1000g of water. Calculate the vapour pressure of solution at `20^(@)C`. The vapour pressure of water of 298 K is 0.023 atm. Assume that the solution behaves ideally.
(Molar mass of sucrose `= 342 g mol^(-1)`).

Explanation

Given, `W_(B) = 68.4 g`
`therefore` Number of moles of sucrose `= (68.4)/(342)`
`W_(A) = 1000g`
`therefore` Number of moles of water `= (1000)/(18)`
According to lowering of vapour pressure,
`(p^(@) - p)/(p^(@)) = chi_(2)`
where, `p^(@)` = vapour pressure of pure water = 0.023 atm p = vapour pressure of solution = ?
`therefore (0.023 - p)/(0.023) = ((68.4)/(342))/((68.4)/(342) + (1000)/(18))`
`(0.023 - p)/(0.023) = 0.00359`
`0.023 - p = 0.00008257 rArr p = 0.02292` atm.