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

(N/A) Given: Molarity $(M)$ = $3 \ mol \ L^{-1}$,Density $(d)$ = $1.25 \ g \ mL^{-1}$,Molar mass of $NaCl$ = $58.5 \ g \ mol^{-1}$.
Mass of $NaCl$ in $1 \ L$ solution = $3 \ mol \times 58.5 \ g \ mol^{-1} = 175.5 \ g$.
Mass of $1 \ L$ solution = $Volume \times Density = 1000 \ mL \times 1.25 \ g \ mL^{-1} = 1250 \ g$.
Mass of solvent (water) = $Mass \ of \ solution - Mass \ of \ solute = 1250 \ g - 175.5 \ g = 1074.5 \ g = 1.0745 \ kg$.
Molality $(m)$ = $\frac{\text{Moles of solute}}{\text{Mass of solvent in } kg} = \frac{3 \ mol}{1.0745 \ kg} \approx 2.79 \ m$.