`2.82g` of glucose (molar mass `=180`) is dissolved in `30g` of water. Calculate the (i) Molality of the solution (ii) mole fractions of (a) glucose (b) water.

Explanation

(a) Calculation of molality of solution.
Mass of glucose = 2.82 g
Mass of glucose `(2.82)180` (Molar mass =180)
Mass of water = 30g
Molality `=("Moles of glucose")/("Mass of water") xx 1000`
`=(2.82 xx 1000)/(180 xx 30) = 0.522`m.
(b) Calculation of mole fraction
Moles of glucose `=(2.82)/180 = 0.0157`
Moles of water `=30/18 = 1.67`
Mole fraction of glucose `=(0.0157)/(0.0157 + 1.67) = 0.009`
Mole fraction of water `=1.67/(0.0157 + 1.67) = 0.991`