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  • At room temperature, a dilute solution of urea is prepared by dissolving 0.60g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mmHg, lowering of vapour pressure(in  mmHg)  will be: (molar mass of

At room temperature, a dilute solution of urea is prepared by dissolving 0.60g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mmHg, lowering of vapour pressure(in  mmHg)  will be: (molar mass of urea = 60g/mol)
Option: 1 0.017
Option: 2 0.028
Option: 3 0.027
Option: 4 0.031
 

Answers (1)

best_answer

Molecular weight of urea = 60 g / mol

Now, we know that :

Relative lowering of vapour pressure is equal to the molar mass of the solute.

\therefore \Delta P=35\times\frac{(\frac{0.6}{60})}{\frac{360}{18}+\frac{0.6}{60}}

\Delta P=35\times\frac{0.01}{20+0.01}

\Delta P=35\times0.00049

\Delta P=0.0174 \: mm Hg

\therefore option (1) is correct.

 

 

Posted by

Ritika Jonwal

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