Two liquids X and Y forms an ideal solution at 300 K, vapour pressure of the solution containing 1 mole of X and 3 moles of Y is 550 mm Hg. At the same temperature, 1 mole of Y is further added to this solution vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mm Hg) of X and Y in their pure states will be, respectively.

200 and 300

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300 and 400

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400 and 600

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500 and 600

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Solution

The correct option is C 400 and 600
$550 = (\frac{1}{4}) (P_{x}^{0}) + (\frac{3}{4}) (P_{y}^{0}) 2200 = P_{x}^{\circ} + 3 P_{y}^{\circ} . . . . . . (A) 560 = (\frac{1}{5}) (P_{x}^{\circ}) + (\frac{4}{5}) (P_{y}^{\circ}) 2800 = P_{x}^{\circ} + 4 P_{y}^{\circ} . . . . . . . . (B)$
On solving A and B, we get,
$P_{x}^{\circ} = 400 a n d P_{y}^{\circ} = 600$

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Similar questions

Q. Two liquids

X

and

Y

form an ideal solution. At

300 K

, vapour pressure of the solution containing

1 m o l

X

and

3 m o l

Y

550 m m H g

. At the same temperature, if

1 m o l

Y

is further added to this solution, vapour pressure of the solution increases by

10 m m H g

. Vapour pressure (in

m m H g

) of

X

and

Y

in their pure states will be ______ respectively.

Two liquids X and Y form an ideal solution.At 300 K,vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mm Hg.At the same temperature, if 1 mol of Y is further added to this solution,vapour pressure of the solution increases by 10mm Hg.So, vapour pressure (in mm Hg) of X and Y in their pure states will be, respectively,

Two liquids $X$ and $Y$ form an ideal solution at 300 K, vapour pressure of the solution containing 1 mole of $X$ and 3 mole of $Y$ is 550 mm Hg. At the same temperature, 1 mole of $Y$ is further added to this solution and the vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mm Hg) of $X$ and $Y$ in their pure states will be, respectively: