Question

In a solution, the concentration of $CaC{l}_2$ is 5 M and concentration of $MgC{l}_2$ is 5 m. If the specific gravity of the solution is 1.05, the concentration of Cl- in the solution is

• A. 10 M
• B. 20 M
• C. 18.5 M
• D. 17.12 M

Answer 1

Answer: (D)

$\frac{1}{M}=\frac{1}{\rho }\left(\frac{1}{m}+10^{-3}\times M^{\prime }\right)$
$M^{\prime }=24+71=95;\rho =1.05$
$\therefore \frac{1}{M}=\frac{1}{1.05}\left(\frac{1}{5}+0.095\right)$
or $M=3.56$ or $\frac{1}{M}=0.28095$
$\therefore$ Molarity of $MgC{l}_2=3.56$
$\therefore$ Concentration of $C{l}^{-}=5\times 2+3.56\times 2=17.12M$