Chemistry - Liquid Solution Question with Solution | TestHub

Given, molarity $=2\mathrm{M}$ [i.e 2 mole $\mathrm{NaOH}$ in $1\mathrm{L}$ solution] Density $(\mathrm{d})=1.28{\mathrm{cm}}^{-3}$ Molecular weight of $\mathrm{NaOH}=40{\mathrm{gmol}}^{-1}$ We know that, $$\begin{array}{rl}& \text{ Density }=\frac{\text{ mass of solution }}{\text{ volume of solution }}\\ & \therefore \text{ Mass of solution }=1.28\mathrm{g}{\mathrm{cm}}^{-3}\times 1000\mathrm{mL}=1280\mathrm{g}\\ & \text{ Moreover, molarity }\\ & =\frac{\text{ number of moles of solute }}{\text{ volume of solution }(\text{ in }\mathrm{mL})}\times 1000\\ & \therefore 2=\frac{\text{ mass of solute }}{40\times 1000}\times 1000=80\mathrm{g}\\ & \text{ Now, mass of solution }=\text{ mass of solvent }+\text{ mass of }\\ & \text{ solute }\\ & 1280\mathrm{g}=\text{ mass of solvent }+80\\ & \therefore \text{ Mass of solvent }=1280-80=1200\mathrm{g}=1.2\mathrm{kg}\\ & \text{ Now, molality }=\frac{\text{ number of moles of solute }}{\text{ mass of solvent }(\text{ in kg) }}\\ & =\frac{2}{1.2}=\frac{20}{12}=\frac{5}{3}=1.67\mathrm{m}\\ & \end{array}$$ Alternative method Molality $(\mathrm{m})=\frac{\mathrm{M}}{1000\mathrm{d}-\mathrm{M}\omega \mathrm{t}}\times 1000$ where, $M=$ molarity, $d=$ density of solution $$\mathrm{M}\omega \mathrm{t}=\text{ molar mass of solute. }$$ On substing the given values, we get $$\begin{array}{rl}\text{ Molality }& =\frac{2}{1000\times 1.28-40}\times 1000\\ & =\frac{2}{1280-40}\times 1000=\frac{2000}{1240}=1.612\mathrm{m}\end{array}$$