Question

Calculate the molarity of a $3.3$ molal solution of $\mathrm{KCl}$ whose density is $1.28$$\mathrm{g}/\mathrm{mL}$.

• A. $3.7\mathrm{M}$
• B. $5.0\mathrm{M}$
• C. $3.4\mathrm{M}$
• D. $2.5\mathrm{M}$

Answer 1

The correct option is C

$3.4\mathrm{M}$

The above question can be solved as:

Step 1:

Here given, molality($\mathrm{m}$)$=3.3\left(m\right)$,

density($\mathrm{d}$)$=1.28g/ml$

Step 2:

We have to find molarity($\mathrm{M}$) form the formula:

$\mathrm{m}=\frac{1000\mathrm{M}}{1000\mathrm{d}-{\mathrm{MM}}_{\mathrm{solute}}}$

where, $\mathrm{m}$ is molality of the solution

$\mathrm{M}$ is molarity of the solution

${\mathrm{M}}_{\mathrm{solute}}$ is molar mass of the solute

${\mathrm{M}}_{KCl}=74.5\mathrm{g}/\mathrm{mol}$

Step 3:

Putting all the values for given $\mathrm{KCl}$ solution in above formula:

or, $3.3=\frac{1000\mathrm{M}}{(1000\times 1.28)-(\mathrm{M}\times 74.5)}$

or, $4224=1000\mathrm{M}+245.85\mathrm{M}$

On solving,

$$\begin{gathered} \mathrm{M}=3.39 \\ \mathrm{M}\approx 3.4\mathrm{M} \end{gathered}$$

Therefore, molarity of $\mathrm{KCl}$ solution evaluated to be $3.4\mathrm{M}$

So, option (C) is correct.