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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
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The vapour pressure of a solution having 2.0g of a solute X(molar mass $32gmol^{-1}$) in 100g of $CS_2$ (vapour pressure 854 Torr) is 848.9 Torr.The molecular formula of the solute is

$\begin{array}{1 1}X\\X_2\\X_4\\X_8\end{array} $

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Answer : $X_8$
Mole fraction of solvent =$\large\frac{(100g/76gmol^{-1})}{(100g/76gmol^{-1})+(2g/M)}$
Hence $848.9$ Torr=$\large\frac{(100/76mol^{-1})}{(100/76mol^{-1})+(2g/M)}$(854 Torr)
Solving for M,we get $M=\large\frac{0.994\times 2g}{1.316(1-0.994)mol}$$=251.8gmol^{-1}$
Number of S atoms in a molecule of solute =$\large\frac{251.8amu}{32amu}$$=8$
answered Jul 22, 2014 by sreemathi.v
 

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