Question

A 100 ml of 1 M $NaCl$ solution, 100 ml of 2 M ${MgCl}_2$ solution and 300 ml of 4 M $Mg({NO}_3{)}_2$ solution are mixed and is diluted to $2$. Which of the following is the correct regarding final concentration?

• A. $\left[{Na}^{+}\right]=0.2M$
• B. $\left[{Mg}^{2+}\right]=0.7M$
• C. $\left[C{l}^{-}\right]=0.2M$
• D. $\left[{NO}_3^{-}\right]=1.2M$

Answer 1

Answer: (B), (D)

The correct options are
B $\left[{Mg}^{2+}\right]=0.7M$
D $\left[{NO}_3^{-}\right]=1.2M$

Moles of NaCl = $1\times \frac{100}{1000}$

$\Rightarrow$ 0.1 mole

$NaCl\to N{a}^{+}+C{l}^{-}$

0.1 0.1

Moles of $MgC{l}_2$ = $2\times \frac{100}{1000}$

$\Rightarrow$ 0.2 mole

$MgC{l}_2\to M{g}^{+2}+2C{l}^{-}$

0.2 0.4

Moles of $Mg(N{O}_3{)}_2$ = $4\times \frac{300}{1000}$

​ $\Rightarrow$ 1.2 mole

$Mg(N{O}_3{)}_2\to M{g}^{2+}+2N{O}_3^{-}$

1.2 2.4

A) $\left[N{a}^{+}\right]=\frac{n}{V}=\frac{0.1}{2}=0.05M$

B) $\left[M{g}^{2+}\right]=\frac{n}{V}=\frac{0.2+1.2}{2}=0.7M$

C) $\left[C{l}^{-}\right]=\frac{n}{V}=\frac{0.1+0.4}{2}=0.25M$

D) $\left[N{O}_3^{-}\right]=\frac{n}{V}=\frac{2.4}{2}=1.2M$

Hence , option (B) and (D) are correct .