A 15 % (m/m) aqueous solution of glucose is prepared by dissolving 13.5 g
of glucose in x g of water. Calculate the value of x
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Answer:
15% glucose C
6
H
12
O
6
in H
2
O
15g of glucose in 85g of H
2
O
Solute⟶glucose⟶15g
Solvent⟶H
2
O⟶85g
No. of moles of solute=
MolarMass
GivenMass
Molar Mass of C
6
H
12
O
6
= 6x12+12x1+6x16
= 72+12+96=180
n=
180
15
=0.0833
Density⟶1.5g/mL
1.5g⟶1mL
1g⟶
1.5
1
mL
100g⟶
1.5
1
×100mL
1.5
100
mLofsolution
1.5
100
×
1000
1
Lofsolution
15
1
Lofsolution
Molarity=
Volume of solution in L
Moles of solute
=
1/15
15/180
=
180
15
×
1
15
=1.25M
Molality=
Mass of solvent in kg
Moles of solute
=
85/1000
15/180
=
180
15
×
85
1000
=
51
50
=0.980m
Molarity of solution=1.25M
Molality of solution=0.980m
Explanation:
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