Find the molarity of 9% (w/v) aqueous
solution of glucose. If the density of
the solution is 1.2 g/cm3 then find
its molality and mole fraction.
Answers
Explanation:
10% w/w solution of glucose in water means that 10 g of glucose in present in 100 g of the solution i.e., 10 g of glucose is present in (100 - 10) g = 90 g of water.
Molar mass of glucose (C6H12O6) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol - 1
Then, number of moles of glucose = 10 / 180 mol
= 0.056 mol
∴ Molality of solution = 0.056 mol / 0.09kg = 0.62 m
Number of moles of water = 90g / 18g mol-1 = 5 mol
Mole fraction of glucose (xg) = 0.056 / ( 0.056+5) = 0.011
And, mole fraction of water xw = 1 - xg
= 1 - 0.011 = 0.989
If the density of the solution is 1.2 g mL - 1, then the volume of the 100 g solution can be given as:
= 100g / 1.2g mL-1
= 83.33 mL
=83.33 x 10-3 L
∴ Molarity of the solution = 0.056 mol / 83.33 x 10-3 L
= 0.67 M
Answer:
Explanation:
MolaLity =
10.74
m
Molarity =
6.76
M
mole fraction =
0.162
Explanation:
40% ethylene glycol means
mass of ethylene glycol =
40 g
mass of water (solvent) =
60 g = 0.060 kg
mass of solution =
100 g
1. Molality
Molar mass of ethylene glycol =
62.07 g/mL
no. of moles =
4 g
62.07 g.mol
−
1
=
0.6444 mol
Molality =
no. of moles
mass of solvent in kg
Molality =
0.6444 mol
0.060 kg
=
10.7 m
2. Molarity
Molarity
=
no. of moles
volume of solution in litres
Density
=
1.05 g/cm
3
Volume
=
Mass
Density
Volume
=
100 g
1.05 g/mL
=
95.24 mL
=
0.0952 L
Molarity
=
0.6444 mol
0.0952 L
=
6.76 M or 6.76 mol/L
3. Mole fraction
no. of moles of water
=
60 g
18.02 g.mol
-1
=
3.329 mol
Total moles
=
(3.329 + 0.6444) mol
=
3.974 mol
mole fraction of ethylene glycol
=
0.6444
3.974
=
0.162