Question

The density of 3M solution of $NaCl$ is $1.25g{mL}^{-1}$. Calculate molarity of the solution.

Answer 1

Density= $1.25gm{L}^{-1}$

$1$ Litre of solution contains $1.25gm$ of compound

$1gm$ of compound= $\frac{1}{1.25}L$ volume

$\Rightarrow 1000gm=1kg=\frac{1000}{1.25}L$ of solution.

Molality is defined as number of moles of compound present in $1$kg of solvent.

Thus, given Molality= $3MNaCl$

i.e. $3$ moles are present in $1kg$ of solution.

$3$ moles are present in $\frac{1000}{1.25}L$ of solution.

Molarity is defined as number of moles per $L$ of solution.

If $\frac{1000}{1.25}L$ contain $3$ moles of $NaCl$

Then $1L$ contain $\frac{3\times 1.25}{1000}$ moles of $NaCl$.

Thus, Molarity will be equal to $3.75\times {10}^{-3}$ .