Question

25%w/v glucose have 2.89 gm/ml calculate its molality​

Answer 1

Solution

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