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Q. In a solution, the concentration of $CaCl_2$ is 5 M and concentration of $MgCl_2$ is 5 m. If the specific gravity of the solution is 1.05, the concentration of Cl- in the solution is

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Solution:

$\frac{1}{M}=\frac{1}{\rho}\left(\frac{1}{m}+10^{-3}\times M '\right)$
$M '=24+71=95; \rho=1.05$
$\therefore \frac{1}{M}=\frac{1}{1.05}\left(\frac{1}{5}+0.095\right)$
or $M=3.56$ or $\frac{1}{M}=0.28095$
$\therefore $ Molarity of $MgCl_{2}=3.56$
$\therefore $ Concentration of $Cl^{-}=5\times2+3.56\times2=17.12\,M$