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What weight of solute (mol. Wt. 60) is r...

What weight of solute (mol. Wt. 60) is required to dissolve in 180 g of water to reduce the vapour pressure to `4//5^(th)` of pure water ?

A

120 g

B

150 g

C

200 g

D

60 g

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The correct Answer is:
B
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What weight of solte (mol.wt.60) is required to dissolve in 180g of water to reduce the vapour pressure to 4//5th of pure water?

Vapour pressure of a solvent is the pressure exterted by vapour when they are in equilibrium with its solvent at that temperature. The vapour pressure of solvent is dependent of nature of solvent, temperature, addition of non-volatile solute as well as nature of solute to dissociate or associate. The vapour pressure of a mixture obtained by mixing two valatile liquids is given by P_(M) = P_(A)^(@).X_(A)+P_(B)^(@).X_(B) where P_(A)^(@) and P_(B)^(@) are vapour pressures of pure components A and B and X_(A), X_(B) are their mole fractions in mixture. For solute-solvent system, the relatio becomes P_(M) = P_(A)^(@).X_(A) where B is non-volatile solute. The amount of solute ("mol. wt. 60") required to dissolve in 180 g water to reduce the vapour pressure to 4//5 of the pure water:

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