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

Options

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

Answer

A

Explanation

Density (d) `= ("Mass of solution")/("Volume of solution")`
Mass of solution `= (1.25 gL^(-1)) xx(1000 mL) = 1250 g`
Mass of solvent = Mass of solution - Mass of solute
`= (1250 - 3 xx 58.5) = 1074.30 g`
Molality of solution `= ("No. of moles of solute")/("Mass of solvent in kg")=(("3 mol"))/(("1.074 kg"))`
`=2.79"mol kg"^(-1)=2.79` m.