The density of 2 M aqueous solution of NaOH is 1.28 ~g / cm^3. The molality of the solution is [Given that molecular mass of . NaOH=40 ~g ~mol^-1 ]

Given, molarity $=2 \mathrm{M}$=2M [i.e 2 mole $\mathrm{NaOH}$NaOH in $1 \mathrm{~L}$1 L solution] Density $(\mathrm{d})=1.28 \mathrm{~cm}^{-3}$(d)=1.28 cm−3 Molecular weight of $\mathrm{NaOH}=40 \mathrm{gmol}^{-1}$NaOH=40gmol−1 We know that, $$\begin{aligned} & \text { Density }=\frac{\text { mass of solution }}{\text { volume of solution }} \\ & \therefore \text { Mass of solution }=1.28 \mathrm{~g} \mathrm{~cm}^{-3} \times 1000 \mathrm{~mL}=1280 \mathrm{~g} \\ & \text { Moreover, molarity } \\ & =\frac{\text { number of moles of solute }}{\text { volume of solution }(\text { in } \mathrm{mL})} \times 1000 \\ & \therefore \quad 2=\frac{\text { mass of solute }}{40 \times 1000} \times 1000=80 \mathrm{~g} \\ & \text { Now, mass of solution }=\text { mass of solvent }+ \text { mass of } \\ & \text { solute } \\ & 1280 \mathrm{~g}=\text { mass of solvent }+80 \\ & \therefore \text { Mass of solvent }=1280-80=1200 \mathrm{~g}=1.2 \mathrm{~kg} \\ & \text { Now, molality }=\frac{\text { number of moles of solute }}{\text { mass of solvent }(\text { in kg) }} \\ & =\frac{2}{1.2}=\frac{20}{12}=\frac{5}{3}=1.67 \mathrm{~m} \\ & \end{aligned}$$​ Density = volume of solution  mass of solution ​∴ Mass of solution =1.28 g cm−3×1000 mL=1280 g Moreover, molarity = volume of solution ( in mL) number of moles of solute ​×1000∴2=40×1000 mass of solute ​×1000=80 g Now, mass of solution = mass of solvent + mass of  solute 1280 g= mass of solvent +80∴ Mass of solvent =1280−80=1200 g=1.2 kg Now, molality = mass of solvent ( in kg)  number of moles of solute ​=1.22​=1220​=35​=1.67 m​ Alternative method Molality $(\mathrm{m})=\frac{\mathrm{M}}{1000 \mathrm{~d}-\mathrm{M} \omega \mathrm{t}} \times 1000$(m)=1000 d−MωtM​×1000 where, $M=$M= molarity, $d=$d= density of solution $$\mathrm{M} \omega \mathrm{t}=\text { molar mass of solute. }$$Mωt= molar mass of solute.  On substing the given values, we get $$\begin{aligned} \text { Molality } & =\frac{2}{1000 \times 1.28-40} \times 1000 \\ & =\frac{2}{1280-40} \times 1000=\frac{2000}{1240}=1.612 \mathrm{~m} \end{aligned}$$ Molality ​=1000×1.28−402​×1000=1280−402​×1000=12402000​=1.612 m​