• # question_answer 8) The density of 3M solution of $\text{NaCl}$ is $1.25g\,m{{L}^{-1}}$Calculate the molality of solution.

Answer:

Let volume of solution = 1 litre, i.e., 1000 mL Mass of solution $=1000\times 1.25=1250\text{ }g$ Mass of solute dissolved $=3\times 58.5=175.5\text{ }g$ Mass of solvent $=1250-175.5=1074.5\text{ }g$ Molality $(m)=\frac{{{w}_{B}}\times 1000}{{{m}_{B}}\times {{w}_{A}}}$ $=\frac{175.5\times 1000}{58.5\times 1074.5}=2.79$

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