Question

A solution is made using 30.3 percent by mass CH₂Cl₂ in CHCl₃. At 30 °C, the vapor pressure of pure CH₂Cl₂ is 490 mm Hg, and the vapor pressure of pure CHCl₃ is 260 mm Hg. The normal boiling point of CHCl₃ is 61.7 °C.

A.) What is the mole fraction of CH₂Cl₂ in the solution?

B.) What is the vapor pressure of the solution?

C.) What is the molality of CH2CL2 in the solution?

D.) What is the boiling point of the solution? (Kb= 3.67 celsius /m)

Answer 1

Ans :-

A). Let mass of CHCl₃ i.e. solvent = 100 g

So,

Mass of CH₂Cl₂ = 30.3 g

No. of moles of CHCl₃ (n_CHCl3) = Given mass of CHCl₃ / Gram molar mass of CHCl₃ = 100 g / 119.38 g/mol = 0.83766 mol

and

No.of moles of CH₂Cl₂ (n_CH2Cl2) = Given mass of CH₂Cl₂​​​​​​​ / Gram molar mass of CH₂Cl₂ = 30.3 g/84.93 g/mol = 0.35676 mol

Therefore,

Mole fraction of CH₂Cl₂ = n_CH2Cl2/(n_CHCl3 + n_CH2Cl2)

= 0.35676 mol / (0.35676 mol + 0.83766 mol)

= 0.35676 mol / (1.19442 mol)

= 0.30

Therefore, Mole fraction of CH₂Cl₂ = 0.30

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B). Mole fraction of CHCl₃ = 1 - 0.30 = 0.70

Partial vapor pressure of CHCl₃ (P_CHCl3) = P⁰_CHCl3 x X_CHCl3

= 260 mm Hg x 0.70

= 182 mm Hg

Similarly,

Partial vapor pressure of CH₂Cl₂ (P_CH2Cl2) = P⁰_CH2Cl2 x X_CH2Cl2

= 490 mm Hg x 0.30

= 147 mm Hg

So, vapor pressure of the solution = P_CHCl3 + P_CH2Cl2 = 182 mm Hg + 147 mm Hg = 329 mm Hg

Therefore, vapor pressure of the solution = 329 mm Hg

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C). Molality of CH₂Cl₂ (m) = No. of moles of CH₂Cl₂​​​​​​​/ Mass of solvent CHCl₃ in kg

= 0.35676 mol / 0.100 Kg

= 3.5676 mol/kg

Therefore, Molality of CH₂Cl₂ solution = 3.57 m

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D). T_b - T⁰_b = k_b.m

T_b - 61.7 °C = (3.67 °C/m).(3.57 m)

T_b - 61.7 °C = (3.67 °C/m).(3.57 m)

T_b - 61.7 °C = 13.10 °C

T_b  = 13.10 °C + 61.7 °C

T_b  = 74.8 °C

Therefore, boiling point of the solution = 74.8 °C