The vapour pressure of pure benzene is 640mm og Hg. 2.175×10^{-3}kg of non-vloatile solute is added to 39 gram of benzene the vapour pressure of solution is 600mm of HG. Calculate molar mass of solute.

[C = 12, H = 1]

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#### Solution

Molar mass of benzene C_{6}H_{6} = (6 × 12 + 6 × 1) × 10^{-3} kg mol^{-1}

p_{1}^{0} = 640mm Hg, p = 600mm Hg

W_{1} = 39 × 10^{-3} kg

W_{2} = 2.175 × 10^{-3} kg mol^{-1}

M_{1} = 78 × 10^{-3} kg mol^{-1}

M_{2} = ?

`(p_1^0-p)/p_1^0 = W_2/M_2 M_1/W_1`

`(640mm-600mm)/(640mm) = (2.175xx10^-3"kg"xx78.0xx10^-3"kg mol"^-1)/(39.0xx10^-3"kg"xxM_2)`

`M_2 = (2.175xx10^-3"kg"xx78.0xx10^-3"kg mol"^-1xx640"mm")/(39.0xx10^-3"kg"xx40"mm")`

`M_2 = 69.6xx10^-3"kg mol"^-1`

`"Molecular mass" = 69.6"g mol"^-1`

Concept: Colligative Properties and Determination of Molar Mass - Introduction

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