Question

A $100$ solution of ${Na}_2{CO}_3$ is prepared by dissolved $8.653$ of the salt in water. The density of the solution is $1.0816/$. What are the Molarity and Molality of the solution?
(Atomic mass of Na is 23, e is 12 and of O is 14)

Answer 1

As we know that,

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$

Given:-

Density of solution $=1.0816gm/mL$

Volume of solution $=100mL$

$\therefore 1.0816=\frac{\text{Mass of solution}}{100}$

$\Rightarrow \text{Mass of solution}=108.16g$

Mass of solute $\left({Na}_{2}C{O}_3\right)=8.653g$

Therefore,

Mass of solvent $=108.16-8.653=99.507g$

Molecular mass of ${Na}_{2}C{O}_3=83g$

As we know that,

No. of moles $=\frac{\text{Given mass}}{\text{Mol. mass}}$

$\therefore$ No. of moles of ${Na}_{2}C{O}_3=\frac{8.653}{83}=0.104\text{mol}$

Now,

As we know that,

Molarity $=\frac{\text{No. of moles of solute}}{{\text{Volume of solution}}_{\left(\text{in L}\right)}}$

Therefore,

Molarity $=\frac{0.104}{100}\times 1000=1.04M$

Again, as we know that,

Molality $=\frac{\text{No. of moles of solute}}{{\text{Mass of solution}}_{\left(\text{in Kg}\right)}}$

Therefore,

Molality $=\frac{0.104}{99.507}\times 1000=1.045m$

Hence the molarity and molality of solution are $1.04M$ and $1.045m$ respectively.