Calculate the molality of 1M solution of sodium nitrate. The density of solution is `1.25 "g cm"^(-3)`

Answer

0.86 m

Explanation

`1 M NaNO_(3)` solution contains 85 g of `NaNO_(3)` in 1000 mL of the solution
Density of solution `= 1.25 g mL^(-1)`. Therefore, mass solution `=1000 xx 1.25 = 1250 g`
Mass of solvent (water) `=1250 - 85 = 1165 g = 1.165 kg`
Molality of solution `(m)=(("1 mol"))/(("1.165 kg"))="0.86 mol kg"^(-1)=0.86m`.