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What is molality of 1 M solution of sodium nitrate if its density is `1.25 "g cm"^(-3)` ? |
Answer» Volume of solution `= 1000 cm^(3)` Density of solution `= 1.25 "g cm"^(-3)` Mass of `1000 "cm"^(3)` of solution `= ("1000 cm"^(3))xx("1.25 g cm"^(-3))=1250 g` Mass of `NaNO_(3)` (molar mass ) present `= 23+14 + 48 = 85` g Mass of solvent (water) `= (1250 - 85)=1165 g = 1.165 kg` Molality of solution `(m)=("Mass of "NaNO_(3)//"Molar mass")/("Mass of solvent in kg")` `=((85g)//(85 "g mol"^(-1)))/((1.165 kg))=0.858"mol kg"^(-1)=0.858` m. |
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