`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 soluiton.
Mass of glucose = 2.82 g .
Moles of glucose `= 2.82 g`
Moles of glucose `= (2.82)/(180)` (Molar mass = 180)
Mass of water = 30 g
Molality ` = ("Moles of glucose")/("Mass of water") xx 1000`
` = (2.82 xx 1000)/(180 xx 30) = 0.522 m`
(b) Calculation of mole fraction .
Mole 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`