Find out the mass percentage of solvent in 2 kg of solution of glucose in water which is 25 % (w/w) in terms of solute.
Answers
Answer:
karan aujla song dhol and sidhu moose wala vs
Answer:
10% w/w glucose solution corresponds to 10 g glucose present in 100 g of solution which contains 90 g of water.
(i) Molality is the number of moles of glucose present in 1 kg of water. Number of moles of glucose is the ratio of mass (10 g) of glucose to its molar mass (180 g/mol).
Molality =
180×0.090
10
=0.618 m
Note: 90 g corresponds to 0.090 lg.
(ii) Molarity is the ratio of number of moles of glucose to volume of solution (in L).
Since, denstiy of solution is 1.20 g/mL, the volume of solution is
1.20
100
mL or
1.20×1000
100
L.
Molarity =
180×
1.20×1000
100
10
=0.667 M
(iii) Number of moles of glucose =
180
10
=0.0556
Number of moles of water =
18
90
=5.0
Mole fraction of glucose =
0.0556+5.0
0.0556
=0.011
Mole fraction of water =1−0.011=0.989