1.

The vapour pressure of pure benzene is 640 mm of Hg. `2.175 xx 10^(-3)` kg of non-volatile solute is added to 39 gram of benzene, the vapour pressure of solution is 600 mm of Hg. Calculate molar mass of solute (C = 12, H = 1).

Answer» Molar mass of benzene
`= (6 xx 12 + 6 xx 1)`
` 78 xx 10^(-3) kg mol^(-1)`
`P_(0) = 640 mm Hg,`
P = 600 mm Hg
`W_(1) = 39 xx 1-^(-13) kg`
`W_(2) = 2.175 xx 10^(-3) kg`
`M_(1) = 78 xx 10^(-3) kg, M_(2) = ?`
`M_(2) = (p_(0) xx W_(2) xx M_(1))/(Delta P xx W_(1))`
`M_(2) = (640 xx 2.175 xx 10^(-3) xx 78 xx 10^(-3))/((640 - 600) xx 39 xx 10^(-3))`
`= 69.6 10^(-3) kg mol^(-1)`


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