Question

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

• A. $3.7M$
• B. $3.4M$
• C. $5.0M$
• D. $2.5M$

Answer 1

The correct option is B $3.4M$

Given,
$\text{Molality = 3.3 m}\text{Density}=1.28\text{g/mL}$
By using formula,
$\text{Molality}=\frac{M\times 1000}{1000d-M\times M_{solute}}$
$\text{3.3}=\frac{M\times 1000}{1000\times 1.28-M\times 74.5}$
$4224-245.85M=1000M$
$1245.85M=4224$
$M=\frac{4224}{1245.85}$
$M=3.39M\approx 3.4M$