Question

vapour pressure of pure A is 10 torr, when 1 gm B is dissolved in 20 gm A, its vapour pressure reduced upto 9 torr.
molecular mass of A is 200 amu then for B it will be-
100 a.m.u.
2) 90 a.m.u.
3) 75 a.m.u.
4) 120 a.m.u.
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Answer 1

i is the answer is coming to the same

Answer 2

Given :-

• The  vapour pressure of pure A is 10 torr and at the same temperature when 1 g of B is dissolved in 20 gm of A, its vapour pressure is reduced to 9.0 torr. If the molecular mass  of A is 200 amu,

To Find :-

• Molecular mass of B

Solution :-

$\sf \dfrac{10-9}{10} = \dfrac{\dfrac{1}{x}}{\dfrac{20}{200}+\dfrac{1}{x}}$

$\sf \dfrac{1}{10} = \dfrac{\dfrac{1}{x}}{\dfrac{20}{200}+\dfrac{1}{x}}$

$\sf \sf \dfrac{1}{10} = \dfrac{\dfrac{1}{x}}{\dfrac{1}{10}+\dfrac{1}{x}}$

$\sf \dfrac{1}{10}\bigg(\dfrac{1}{10} + \dfrac{1}{x}\bigg) = \dfrac{1}{x}$

$\sf \dfrac{1}{100} + \dfrac{1}{10x} = \dfrac{1}{x}$

$\sf \dfrac{1}{100} = \dfrac{1}{x}- \dfrac{1}{10x}$

$\sf \dfrac{1}{100} = \dfrac{10x-x}{10}$

$\sf 10= 100(9x)$

$\sf x = \dfrac{900}{10} = 90$