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

Options

A. The percentage by mass of `NaNO_(3) = 6.8`
B. The percentage by mass of `H_(2)O = 93.2`
C. The molality of the solution is `10.72`
D. The solution has `0.2` moles of `NaNO_(3)`

Answer

A, B, C

Explanation

Molarity ` = (10 xx d xx x)/M_(o) `
where d = density of solution, `x = % ` by mass of `NaNO_(3), M_(o) = " mol wt. of " NaNO_(3) = 85`
`1 = (10 xx 1.25 xx (x))/85 rArr = 85/(10 xx 1.25) = 6.8`
` :. % " by mass of " H_(2)O = 100 - 6.8 = 93.2`
Hence , 1 mole of `NaNO_(3)` is present in `93.2 " g of " H_(2)O`
` :. ` Molality ` = 1/(93.2) xx 1000 = 10.72 ` molal