A solution of acetic acid has molarity equal to $1.35\mathrm{M}$ and molarity equal to $1.45\mathrm{mol}{\mathrm{kg}}^{-1}$. The density of solution will be 

For $1.35\mathrm{M}$ solution. we will have $n_2=1.35\mathrm{mol}$ and $V=1000\mathrm{mL}$

The solution is also $1.45\mathrm{mol} {\mathrm{kg}}^{-1}$. The mass of solvent to have $1.35\mathrm{mol}$ of solute will be 

$m_1=\left(\frac{1.35\mathrm{mol}}{1.45\mathrm{molkg}}\right)=0.93103\mathrm{kg}=931.03\mathrm{g}$

The mass of solution will be $m=m_1+n_2{M}_2=931.03\mathrm{g}+\left(1.35\mathrm{mol}\right)\left(60\mathrm{g}{\mathrm{mol}}^{-1}\right)=1012.03\mathrm{g}$

The density of the solution will be $\rho =\frac{m}{V}=\frac{1012.03\mathrm{g}}{1000\mathrm{mL}}=1.012\mathrm{g}{\mathrm{mL}}^{-1}$.