Two liquids A and B form an ideal solution. At 300 K the vapour pressure of a solution of 1 mole of A and x moles of B is 550 mm of Hg. If the vapour pressures of pure A and B are 400 mm of Hg nd 600 mm of Hg respectively, then x is -

Question

Two liquids A and B form an ideal solution. At 300 K the vapour pressure of a solution of 1 mole of A and x moles of B is 550 mm of Hg. If the vapour pressures of pure A and B are 400 mm of Hg nd 600 mm of Hg respectively, then x is - (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

PT=PA° XA+PB° XB 550=400((1/1 + x))+600((x/1 + x)) 55=(40/1 + x)+(60 x/1 + x) 55+55x=40+60x x=3

Q. Two liquids $A$ $an d$ $B$ form an ideal solution. At $300$ $K$ the vapour pressure of a solution of $1$ $m o l e$ of $A$ and $x$ moles of $B$ is $550$ $mm$ of Hg. If the vapour pressures of pure $A$ $an d$ $B$ are $400$ $mm$ of Hg nd $600$ $mm$ of Hg respectively, then $x$ is -

$1$

$2$

$3$

$4$

$P_{T} = P_{A}^{\circ} X_{A} + P_{B}^{\circ} X_{B}$

$550 = 400 (\frac{1}{1 + x}) + 600 (\frac{x}{1 + x})$

$55 = \frac{40}{1 + x} + \frac{60 x}{1 + x}$

$55 + 55 x = 40 + 60 x$

$x = 3$

Q. Two liquids A and B form an ideal solution. At 300 K the vapour pressure of a solution of 1 mole of A and x moles of B is 550 mm of Hg. If the vapour pressures of pure A and B are 400 mm of Hg nd 600 mm of Hg respectively, then x is -

1

2

3

4

PT​=PA∘​XA​+PB∘​XB​ 550=400(1+x1​)+600(1+xx​) 55=1+x40​+1+x60x​ 55+55x=40+60x x=3

Q. Two liquids $A$ $an d$ $B$ form an ideal solution. At $300$ $K$ the vapour pressure of a solution of $1$ $m o l e$ of $A$ and $x$ moles of $B$ is $550$ $mm$ of Hg. If the vapour pressures of pure $A$ $an d$ $B$ are $400$ $mm$ of Hg nd $600$ $mm$ of Hg respectively, then $x$ is -

$1$

$2$

$3$

$4$

$P_{T} = P_{A}^{\circ} X_{A} + P_{B}^{\circ} X_{B}$

$550 = 400 (\frac{1}{1 + x}) + 600 (\frac{x}{1 + x})$

$55 = \frac{40}{1 + x} + \frac{60 x}{1 + x}$

$55 + 55 x = 40 + 60 x$

$x = 3$