Question

If the mass percentage of glucose in aqueous

solution is 18%, then the molality of glucose

in the solution is (Molar mass of glucose

= 180 g/mol)

(1) 1.0 m (2) 3.61 m

(3) 4.52 m (4) 1.22 m​

Answer 1

Answer:

1.22m

Explanation:

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Answer 2

Given :

Mass percentage of Glucose = 18%

To Find :

Molality of glucose in the solution

Solution :

Mass percentage of glucose in the solution is 18%, this means in 100g of solution, 18 g of glucose is present.

No. of moles of glucose = $\frac{Given Mass}{Molar Mass}$  = $\frac{18}{180}$ = 0.1

We know,

Molality = [No. of moles of the solute ÷ weight of the solvent] × 1000

No. of moles of solute = 0.1

Weight of the solvent = (100 - 18) = 82 g

Molality = $\frac{0.1}{82}$ × 1000

             = 1.22 m

∴ The molality of glucose  in the solution is 1.22 m

The correct option is (4) 1.22 m