A solution is prepared by mixing 0.0450 mole of and 0.0600 mole of at 25°C. Assuming the solution is ideal, calculate the composition of the vapor (in terms of mole fractions) at 25°C. At 25°C the vapor pressures of pure and pure are 133 and 11.4 torr, respectively

A solution is prepared by mixing 0.0450 mole of CH₂Cl₂ and 0.0600 mole of CH₂Br₂ at 25°C. Assuming the solution is ideal, calculate the composition of the vapor (in terms of mole fractions) at 25°C. At 25°C, the vapor pressures of pure CH₂Cl₂ and pure CH₂Br₂ are 133 and 11.4 torr, respectively

The answer to the question is

The composition of the vapor in terms of mole fractions = 0.898 of CH₂Cl₂ and 0.102 of CH₂Br₂

Explanation:

To solve this, we calculate the mole fraction as follows

1)

CH₂Cl₂ ⇒ 0.0450 mol / (0.0450 mol + 0.0600 mol) = 0.045/0.105 = 0.429

CH₂Br₂ ⇒ 0.0600 mol / (0.0450 mol + 0.0600 mol) = 0.06/0.105 =0.571

By  Raoult's Law we hve

Pressure of solution = P°Cl χCl + P°Br χBr

Psolution =(0.429) (133 torr)  +(0.571) (11.4 torr)  = 57.057 torr + 6.509 torr = 63.566 torr  

3) Calculate mole fractions in vapor:

CH₂Cl₂ ⇒ 57.057 / 63.566 = 0.898

CH₂Br₂ ⇒ 6.509 / 63.566 = 0.102

The composition of the vapor in terms of mole fractions is 0.898 of CH₂Cl₂ and 0.102 of CH₂Br₂

idk what the question is about