Question

If the mass percentage of glucose in the aqueous
solution is 36% then the molality of glucose in the
solution will be

Answer 1

Answer:

Molalty = 3.125 moles/kg.

Explanation:

Given, mass percentage of glucose = 36%.

Let the mass of the solution be 100g.

then,mass of glucose = 36 g.

Molar mass of glucose = 180g

Hence, number of moles of glucose =

Now, weight if solvent = 64 g =

We know, molality = Hence, molalty = 3.125 moles/kg.

Answer 2

Answer:

The molality of glucose solution is equal to 3.125m.

Explanation:

Given the mass percentage of glucose = 36%

Consider the mass of solution $=100g$

Mass of glucose(solute) $=36g$

Mass of solvent $=100-36=64g$

Molar mass of glucose $M=180gmol^{-1}$

Number of moles of glucose $=\frac{36}{180}=0.2mol$

Molality = [(no. of moles of solute)/(mass of solvent)]×1000

Molality of glucose solution, $m=\frac{0.2}{64}*1000$

Molality, $m=3.125molkg^{-1}$