1.

You are provided with 1 M solution of NaNO_(3) whose density = 1.25g/mL

Answer»

Thepercentage by mass of `NaNO_(3) = 6.8`
The percentage by mass of `H_(2)O = 93.2`
The molality of the SOLUTION is `10.72`
The solution has `0.2` moles of `NaNO_(3)`

Solution :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


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