Question

Two liquids p andq form an ideal solution at 300 k the vapour pressure of a solution of 1 mol of p and x mole of q is 550 mm. If the vapor pressure of pure p and pure q are 400 mm and 600 mm respectively then x is

Answer 1

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Answer 2

Answer:

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Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

$p_A= x_A \times p_A^0$ and $p_B= x_B \times p_B^0$

where, x = mole fraction

$p^0_A$ = pressure in the pure state  of solution A

$p^0_B$ = pressure in the pure state  of solution B

${x_A}$ = mole fraction of solution A

${x_B}$ = mole fraction of solution B

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_A+p_B$

$p_{total}=x_Ap_A^0+x_Bp_B^0$

given $x_{p} = 1 mol$, $x_{q} = x mol$

$p_{p}^0 = 400 mm$

$p_{q}^0 = 600 mm$

$p_{total}$ = 550 mm

$550 mm = 1 mol \times 400 mm + x mol \times 600$

               x = 0.25 mol

The mole fraction of solution q in vapour phase is 0.25 mol.