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  • A set of solutions is prepared using 180 g of water as a solvent and 10 g of different non-volatile solutes A,B and C. The relative lowring of vapour pressure in the presence of these solutes are in the order [Given, molar mass of A=100 g mol -1</sup

A set of solutions is prepared using 180 g of water as a solvent and 10 g of different non-volatile solutes A,B and C. The relative lowring of vapour pressure in the presence of these solutes are in the order [Given, molar mass of A=100 g mol -1; B=200 g mol -1; C=10,000 g mol -1]
Option: 1 B>C>A
Option: 2 C>B>A
Option: 3 A>B>C
Option: 4 A>C>B

Answers (1)

best_answer

Here vapour pressure of water is lowering.

We know this formula, 

Partial Vapour pressure of water Pwater = PoXwater

Lowering Vapour pressure of water = Po – Pwater

The relative lowering of vapour pressure  \mathrm{=\frac{P^o-P}{P^o}}=\mathrm{\frac{\Delta P}{P^o}} =1- \chi_{water}

\mathrm{\mathrm{\frac{\Delta P}{P^o}} =\chi_{solute}=\frac{\frac{m_{solute}}{M_{solute}}}{\frac{m_{solute}}{M_{solute}}+\frac{m_{water}}{M_{water}}}}

Now, 

\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{A}}=\frac{\frac{10}{100}}{\frac{10}{100}+\frac{180}{18}}

\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{B}}=\frac{\frac{10}{200}}{\frac{10}{200}+\frac{180}{18}}

\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{C}}=\frac{\frac{10}{10,000}}{\frac{10}{10,000}+\frac{180}{18}}

From above,

\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{A}}>\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{B}}>\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{C}}

Posted by

Kuldeep Maurya

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