Question

To a pure $100ml$ liquid of ${}^{\prime }{A}^{\prime }$ having molarity $3M$ and density $\frac{3}{2}\text{g/ml}$, $100\text{ml}$ of another pure liquid ${}^{\prime }{B}^{\prime }$ having molarity $8M$ and density $2\text{g/ml}$ is mixed, then find molality of final liquid solution formed.
[Divide your answer by $0.3$]

Answer 1

We know,
Number of moles in $100ml$ of $3M$ = $0.3mol$
Again, density of pure A = $\frac{3}{2}\text{g/ml}$
So, mass of $100ml$ pure A = $\frac{3}{2}\times 100=150g$
So molar mass of A $\left(M_A\right)=\frac{150}{0.3}=500g$

Again,
number of moles of B in $100ml$, $8M$ = $0.8mol$
Given, density of pure B = $2\text{g/ml}$
So, mass of $100ml$ of pure B = $2\times 100g=200g$

So B is the solvent here.
Hence molality of the final solution $=\frac{0.3}{200}\times 1000=1.5m$
So our answer will be $=\frac{1.5}{0.3}=5$