The vapour pressure of pure benzene is 640 mm of Hg. 2.175 x 10-3 kg o

1.	The vapour pressure of pure benzene is 640 mm of Hg. 2.175 x 10-3 kg of non-volatile solute is added to 39 g of benzene, the vapour pressure of solution is 600 mm of Hg. Calculate molar mass of solute (C = 12, H = 1).
Answer» Given: Vapour pressure of pure benzene (\(p^o_1\))= 640 mm Hg Mass of solute (W₂) = 2.175 x 10^-3 kg Mass of solvent (W₁) = 39 g = 39 x 10^-3 kg Vapour pressure of solution (p) = 600 mm Hg To find: Molar mass of solute (M₂) Formula: \(\frac{p^o_1 - p}{p^o_1}=\frac{W_2}{W1}\times\frac{M_1}{M_2}\) Calculation: Molar mass of solvent (benzene) (M₁) = 78 x 10^-3 kg mol^-1 From formula, \(\frac{640-640}{640}=\frac{2.175\times10^{-3}}{39\times10^{-3}}\times\frac{78\times10^{-3}}{M_2}\) M₂ = \(\frac{640\times2.175\times10^{-3}\times78\times^{-3}}{39\times10^{-3}\times40}\) = 69.6 x 10^-3 kg/mol = 69.6 g/mol

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

Answer»

Given:

Vapour pressure of pure benzene (\(p^o_1\))= 640 mm Hg

Mass of solute (W₂) = 2.175 x 10^-3 kg

Mass of solvent (W₁) = 39 g = 39 x 10^-3 kg

Vapour pressure of solution (p) = 600 mm Hg

To find: Molar mass of solute (M₂)

Formula:

\(\frac{p^o_1 - p}{p^o_1}=\frac{W_2}{W1}\times\frac{M_1}{M_2}\)

Calculation: Molar mass of solvent (benzene) (M₁) = 78 x 10^-3 kg mol^-1

From formula,

\(\frac{640-640}{640}=\frac{2.175\times10^{-3}}{39\times10^{-3}}\times\frac{78\times10^{-3}}{M_2}\)

M₂ = \(\frac{640\times2.175\times10^{-3}\times78\times^{-3}}{39\times10^{-3}\times40}\) = 69.6 x 10^-3 kg/mol = 69.6 g/mol

Discussion

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