Question

Mole fraction of glycerine C3H5(OH)3 in solution containing 36 g

of water and 46 g of glycerine is

(a) 0.46

(b) 0.40

(c) 0.20

(d) 0.36​

Answer 1

Answer :-

Mole fraction of glycerine in the solution is 0.20 [Option.(c)]

Explanation :-

We have :-

→ Mass of glycerine = 46 g

→ Mass of water = 36 g

→ Molar mass of glycerine

= 12 × 3 + 1 × 5 + 16 × 3 + 1 × 3

= 36 + 5 + 48 + 3

= 92 g/mol

→ Molar mass of water = 18 g/mol

______________________________

Moles of glycerine [C₃H₅(OH)₃]

= Given mass/Molar mass

= 46/92

= 0.5 mole

Moles of water [H₂O]

= Given Mass/Molar mass

= 36/18

= 2 moles

Mole fraction of glycerine :-

= Moles of glycerine/Total moles

= 0.5/(0.5 + 2)

= 0.5/2.5

= 0.20

Answer 2

Answer:

Option C is correct.

Explanation:

(C × 3) + (H × 5) + (OH × 3) + (16 × 3)

(12 × 3) + (1 × 5) + (1 × 3) + (48)

36 + 5 + 3 + 48

44 + 48

92 gm/mol

Now

Finding moles of water

Moles = Given Mass/Molar mass

Moles = 36/18

Moles = 2

Now

Total moles = 2 + 0.5

Total moles = 2.5

Now

Mole fraction = 0.5/2.5 = 0.20