Molarity and molality of pure ${\mathrm{CH}}_3\mathrm{COOH}$ are respectively ;$\left(d_{{\mathrm{CH}}_3\mathrm{COOH}}=1.5\mathrm{g}/\mathrm{ml}\right)$

$M=\frac{n}{v}$

M = molar concentration 

n = moles of solute 

v = liters of solution

$M=\frac{1}{\frac{60}{1.5}}\times 100$

$m=\frac{mol}{kg}$

m = molality

mol = moles of solute 

{kg} = kilogram of solvent

$$\begin{gathered} m=\frac{1}{60}\times 1000 \\ =16.67 \end{gathered}$$