Name: Solid Solutions
Uploaded: 2019-07-19
Duration: 2 min 56 s
Description: Solutions can be categorized as solid, liquid, and gaseous based on the physical states of solvents and solutes. In this video, let's learn about solid solutions.

Question

An aqueous solution of glucose boils at

100.02 ° C

. What is the number of glucose molecules in the solution containing

100 g

of water? What will be the osmotic pressure of this glucose solution at

27 ° C

? (Given :

K_{b}

for water

0.5 K kg {mol}^{- 1}

,

R = 0.082 L atm {mol}^{- 1} K^{- 1}

, Avogadro's constant,

N_{A} = 6.02 \times 10^{23} {mol}^{- 1}

)

Accepted Answer

Given an aqueous solution of glucose boils at $100.02 ° C$

Temperature $27 + 273 = 300 K$

Elevation in boiling point $K_{b} = 0.5 K kg {mol}^{- 1}$

Universal gas constant $R = 0.082 L atm {mol}^{- 1} K^{- 1}$

Avogadro's constant $N_{A} = 6.02 \times 10^{23} {mol}^{- 1}$

The boiling point of water is $100 ° C$

But, given an aqueous solution of glucose boils at $100.02 ° C$

Thus, elevation in boiling point $∆ T_{b}$ will be $100.02 ° C - 100 ° C = 0.02 ° C$

But, $∆ T_{b} = K_{b} \times m$

On putting the given values we cal calculate the molality of the solution as follows:

$m = \frac{∆ T_{b}}{K_{b}} = \frac{0.02}{0.5} = 0.04$

Thus, molality is $0.04$

We know that molality is given by number of moles of solute and weight of solvent

Weight of the solvent is $100 g = 0.1 kg$

Thus, the number of moles of glucose will be

$n = \frac{0.04}{0.1} = 0.004$

The number of molecules can be obtained by dividing the number of moles of glucose by Avogadro’s number.
Number of molecules of glucose will be

$0.004 \times 6.02 \times 10^{23} = 2.408 \times 10^{21}$ molecules

Osmotic pressure can be obtained by substituting the values of molality, ideal gas constant and temperature in Kelvin.
Thus, osmotic pressure $π = 0.04 \times 0.821 \times 300 = 0.9852 atm$

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