Vapour pressure of pure 'A' is 70 mm of Hg at 25° C. It forms an ideal solution with 'B' in which mole fraction of A is 0.8. If the vapour pressure of the solution is 84 mm of Hg at 25°C, the vapour pressure of pure 'B' at 25°C is

Question

Vapour pressure of pure 'A' is 70 mm of Hg at 25° C. It forms an ideal solution with 'B' in which mole fraction of A is 0.8. If the vapour pressure of the solution is 84 mm of Hg at 25°C, the vapour pressure of pure 'B' at 25°C is (A) 70 mm (B) 140 mm (C) 28 mm (D) 56 mm

Tardigrade Admin · Accepted Answer

p=pA° xA+pB° xB ⇒ 84=70 × 0.8+pB° × 0.2 84=56+pB° × 0.2 pB°=(28/0.2)=140 mm

$70 mm$

$140 mm$

$28 mm$

$56 mm$

$p = p_{A}^{\circ} x_{A} + p_{B}^{\circ} x_{B}$

$\Rightarrow 84 = 70 \times 0.8 + p_{B}^{\circ} \times 0.2$

$84 = 56 + p_{B}^{\circ} \times 0.2$

$p_{B}^{\circ} = \frac{28}{0.2} = 140 mm$

Q. Vapour pressure of pure ′A′ is 70mm of Hg at 25∘C. It forms an ideal solution with ′B′ in which mole fraction of A is 0.8. If the vapour pressure of the solution is 84mm of Hg at 25∘C, the vapour pressure of pure ′B′ at 25∘C is

70mm

140mm

28mm

56mm

p=pA∘​xA​+pB∘​xB​ ⇒84=70×0.8+pB∘​×0.2 84=56+pB∘​×0.2 pB∘​=0.228​=140mm

$70 mm$

$140 mm$

$28 mm$

$56 mm$

$p = p_{A}^{\circ} x_{A} + p_{B}^{\circ} x_{B}$

$\Rightarrow 84 = 70 \times 0.8 + p_{B}^{\circ} \times 0.2$

$84 = 56 + p_{B}^{\circ} \times 0.2$

$p_{B}^{\circ} = \frac{28}{0.2} = 140 mm$