Question:
The mass of glucose that should be dissolved in $50\, g$ of water in order to produce the same lowering of vapour pressure as is produced by dissolving $1\, g$ of urea in the same quantity of water is
The mass of glucose that should be dissolved in $50\, g$ of water in order to produce the same lowering of vapour pressure as is produced by dissolving $1\, g$ of urea in the same quantity of water is
Updated On: May 30, 2022
- 1 g
- 3 g
- 6 g
- 18 g
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The Correct Option is B
Solution and Explanation
$\frac{P-P_{s}}{P}=\frac{w_{1} M_{2}}{w_{2} M_{1}}$
To produce same lowering of vapour pressure, $\frac{P-P_{s}}{P}$ will be same for both cases.
So, $\frac{W_{\text {(Glucose) }} \times 18}{50 \times 180} =\frac{W_{\text {(urea) }} \times 18}{50 \times 60} $
$W_{\text {(Glucose) }} $ = weight of glucose
$W_{\text {(Glucose) }}$ = weight of urea
or $\frac{W_{\text {(Glucose) }} \times 18}{50 \times 180}=\frac{1 \times 18}{50 \times 60}$
$W_{\text {(Glucose) }}=3$
To produce same lowering of vapour pressure, $\frac{P-P_{s}}{P}$ will be same for both cases.
So, $\frac{W_{\text {(Glucose) }} \times 18}{50 \times 180} =\frac{W_{\text {(urea) }} \times 18}{50 \times 60} $
$W_{\text {(Glucose) }} $ = weight of glucose
$W_{\text {(Glucose) }}$ = weight of urea
or $\frac{W_{\text {(Glucose) }} \times 18}{50 \times 180}=\frac{1 \times 18}{50 \times 60}$
$W_{\text {(Glucose) }}=3$
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Concepts Used:
Colligative Properties
Colligative Property of any substance is entirely dependent on the ratio of the number of solute particles to the total number of solvent particles but does not depend on the nature of particles. There are four colligative properties: vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.
Examples of Colligative Properties
We can notice the colligative properties of arrangements by going through the accompanying examples:
- On the off chance that we add a spot of salt to a glass full of water, its freezing temperature is brought down impressively than its normal temperature. On the other hand, the boiling temperature is likewise increased and the arrangement will have a lower vapor pressure. There are also changes observed in its osmotic pressure.
- In the same way, if we add alcohol to water, the solution’s freezing point goes down below the normal temperature that is usually observed for either pure alcohol or water.
Types of Colligative Properties
- Freezing point depression: ΔTf =1000 x kf x m2 /(M2 x m1)
- Boiling point elevation: ΔTb = kb m
- Osmotic pressure: π = (n2/V) RT
- Relative lowering of vapor pressure: (Po - Ps)/Po