Question

calculate the molality of an aqueous solution in which mass percentage of glucose is 30 %

Answer 1

Answer:-

Mass percent of glucose is 30%

means, 30/100(w/w)

Molality = no. of moles of solute / mass of solvent in kg

molar mass of glucose is 180g

Molality= 30×1000/180×100

=1.6m

Answer 2

Given - Mass percentage of glucose : 30%

Find - Molality of aqueous solution.

Solution - Mass percentage of glucose is 30% which states that 30 gram of glucose is present in 100 gram of solution.

Amount of water in solution : 100 - 30 = 70 gram.

Molality is calculated by the formula - number of moles of solute/molecular weight of solvent in kilogram.

Number of moles of glucose = weight/molecular weight.

Number of moles of glucose = 30/180 = 1/6

Keeping the values in equation-

Molality = (1/6)/(70/1000)

Molality = 1000/420 = 2.38 m.

Hence, the Molality of aqueous solution will be 2.38 m.