Question

If the mass percentage of glucose in the aqueous
solution is 36% then the molality of glucose in the
solution will be
(1) 2.1 m
(2) 3.1 m
(3) 4.5 m
(4) 6.2 m​

Answer 1

Heya,

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 = $\frac{36}{180} = \frac{1}{5}$

Now, weight if solvent = 64 g = $64 \times {10}^{ - 3} kg$

We know,

molality = $\frac{no \: of \: moles \: of \: solute}{weight \: of \: solvent \: in \: kg}$

= $\frac{ \frac{1}{5} }{64 \times {10}^{ - 3} } = \frac{1000}{5 \times 64} = \frac{200}{64} \\ = \frac{50}{16} = 3.125$

Hence, molalty = 3.125 moles/kg.

Hope helped !