Question

At 25°C, 5 % aqueous solution of glucose (molecular weight = 180 g mol–1 ) is isotonic with a 2% aqueous solution containing an unknown solute. What is the molecular weight of the unknown solute?

Answer 1

Answer:

$\pi \: {glucose} = \pi \: unknown \: solute$

$\frac{m1}{ M1 } = \frac{m2}{M2}$

$\frac{5}{180} = \frac{2}{ M2 }$

$=> M2 = 72$

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Answer 2

Answer:

Explanation:

At 25°C, 5 % aqueous solution of glucose (molecular weight = 180 g mol–1 ) is isotonic with a 2% aqueous solution containing an unknown solute. What is the molecular weight of the unknown solute.

Solution:

At 25°C, 5 % aqueous solution of glucose,

(molecular weight = 180 g mol–1 ) isotonic with a 2% aqueous solution containing an unknown solute.

Given that,

Since the two solutions are isotonic, they must have same concentrations in moles/litre.

For glucose solution, concentration

= $5g/100cm^{2}$ (given)

= $50g/L$

∴ $\frac{50}{100}$  $=\frac{20}{M}$

or $M=72$

( ∵ Molar mass of glucose = $180 g$ $mol^{-1}$)

For unknown substance, concentration

=  $2g/100cm^{3}$ (given)

= $20g/L$

= $\frac{20}{M} mol/ L$

∴ $\frac{50}{180}=\frac{20}{M}$

or $M=72$

Hence proved

The molecular weight of the unknown solute is $M=72$

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