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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
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The expression relating mole fraction of solute $(x_2)$ and molarity $(M)$ of the solution is (where $\rho$ is the density of solution,$M_1$ and $M_2$ are the molar masses of solvent and solute,respectively)

$\begin{array}{1 1}x_2=\large\frac{MM_1}{M(M_1-M_2)+\rho}\\x_2=\large\frac{MM_1}{M(M_1-M_2)-\rho}\\x_2=\large\frac{M(M_1-M_2)+\rho}{MM_1}\\x_2=\large\frac{M(M_1-M_2)-\rho}{MM_1}\end{array} $

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1 Answer

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Answer : $x_2=\large\frac{MM_1}{M(M_1-M_2)+\rho}$
We have
$x_2=\large\frac{n_2}{n_1+n_2}=\frac{n_2}{(m_1/M_1)+n_2}=\frac{n_2M_1}{m_1+n_2M_1}=\frac{n_2M_1}{(m_1+m_2)+n_2M_1-m_2}$
$\Rightarrow \large\frac{n_2M_1}{V\rho+n_2(M_1-M_2)}=\frac{(n_2/V)M_1}{\rho+(n_2/V)(M_1-M_2)}=\frac{MM_1}{\rho+M(M_1-M_2)}$
answered Jul 17, 2014 by sreemathi.v
 

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