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The density of $3M$ solution of $Na_2S_2O_3$ is $1.25$ $g mL^{−1}$. Calculate the mole fraction of $Na_2S_2O_3$.


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Given, $Na_2S_2O_3$ has molarity = 3 mol $L^{-1}$
Moles of $Na_2S_2O_3$ = 3
$\therefore$ Weight of $Na_2S_2O_3 = 3\times158 = 474$ g
and, Volume of the solution = 1 litre = 1000mL
$\therefore$ Weight of the solution = $1000\times1.25$ = 1250 g
$\therefore$ weight of water = 1250 - 474 = 776 g
Mole fraction of $Na_2S_2O_3 = \large\frac{\text{Moles of }Na_2S_2O_3}{\text{Moles of }Na_2S_2O_3+\text{Moles of }H_2O}$ $=\large\frac{3}{3+\large\frac{776}{18}}$ = 0.065
Hence the answer is (d)
answered Feb 11, 2014 by sharmaaparna1
edited Mar 23, 2014 by balaji.thirumalai

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