The density of `3M` solution of `NaCl` is `1.25 g mL^(-1)`. The molality of the solution is

Options

A. 2.79
B. 1.79
C. 0.79
D. 2.98

Answer

A

Explanation

Molarity = 3M
3 moles of solute present in 1 litre of solution.
`d=(m)/(v)`
Mass of solution = density of solution `xx` volume of solution.
`m=1.25xx1000`
`m=1250g`.
Mass of solvent = mass of solution – mass of solute
`=1250-3xx58.5`
= 1074.50
Molality of solution, `m=(n_("soluble"))/(W_("solvent"(g)))xx1000`
`=(3)/(1074.50)xx1000`
= 2.79 m.