A solution of glucose in water is labelled as 10 % (w/w). The density of the solution is 1.20 g/ml. Calculate the (a) molality (b) molarity   and (c) mole fraction of each component in solution.   

A solution of glucose in water is labelled as 10 % (w/w). The density of the solution is 1.20 g/ml. Calculate the (a) molality (b) molarity   and (c) mole fraction of each component in solution. 

Answer:

Given Values

WGlu = 10 gm

MGlu = C6H12O6 = 6(12)+1(12)+6(16) = 180

WWater = 90 gm

MWater = H2O = 2(1)+16 = 18

DSol = 1.20 gm/ml

WSol = 100 gm

VSol = \( \frac{W_{Sol}}{D_{Sol}} = \frac{100}{1.20} \) = 83.33 ml

\( n_{Glu} = \frac{W_{Glu}}{M_{Glu}} = \frac{10}{180} \) = 0.056

\( n_{Water} = \frac{W_{Water}}{M_{Water}} = \frac{90}{18} \) = 5

Calculations

Calculate molarity of solution:

Molarity = \( \frac{W_{B} \times 1000}{M_{B} \times V_{Sol}} = \frac{10 \times 1000}{180 \times 83.33} \)

Molarity = 0.667 mol/L

Calculate molality of solution:

Molality = \( \frac{W_{B} \times 1000}{M_{B} \times W_{A}} = \frac{10 \times 1000}{180 \times 90} \)

Molality = 0.617 mol/Kg

Calculate mole fraction of Glucose:

χGlu = \( \frac{n_{Glu}}{n_{Glu}+n_{Water}} \)

χGlu = \( \frac{0.056}{0.056+5}  \) = 0.011

Calculate mole fraction of Water:

χWater = 1 – χGlu = 1 – 0.011 = 0.989

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