If 8 g of a non-electrolyte solute is dissolved in 114 g of n-octane to reduce its vapour pressure to 80%, the molar mass (in g mol-1) of the solute is
[Given that the molar mass of n-octane is 114 g mol-1.]
[Given that the molar mass of n-octane is 114 g mol-1.]
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Relative lowering of vapour pressure is a colligative property as it depends on the moles of the solute but is independent of its nature.
- $40$
- $60$
- $80$
- $20$
The Correct Option is A
Approach Solution - 1
Let us take a dilute solution,
\(\frac{P_{0}-P_{s}}{P_{s}} \approx\frac{P_{0}-P_{s}}{P_{0}}=\frac{n_{\text{Solute}}}{n_{\text{Solvent}}}\)
Let \(P_0\) = 100,
Vapour Pressure reduced to 80%
\(\therefore P_{s}=80\)
\(\frac{100-80}{100}=\frac{8 m}{114}\)
\(m=40\)
Therefore, the correct option is (A) : 40.
Approach Solution -2
Relative lowering of vapour pressure is a colligative property as it depends on the moles of the solute but is independent of its nature.
Raoult’s law gives the relation between the relative lowering of the vapour pressure of a solution and the mole fraction of the non-volatile solute by:
\(\frac{p_0-p_s}{p^0}=\frac{n_2}{n_1+n_2}\)
- where p0 denotes the vapour pressure of the pure solvent,
- ps denotes the vapour pressure of the solution,
- n2 denotes the number of moles of the solute, and
- n1 denotes the number of moles of the solvent.
Complete step-by-step solution:
Given: Molar mass of a non-volatile, non-electrolyte solute = 8 grams per mole.
- Mass of octane = 114g
- Also, vapour pressure is reduced to 80 percent.
- The term p0 - ps represents the lowering of vapour pressure
- and \(\frac{p_0-p_s}{p^0}\) represent the relative lowering of vapour pressure
- \(\frac{n_2}{n_1+n_2}\) represent the mole fraction of that solute.
Therefore, we can write: \(\frac{p_0-p_s}{p^0}=\frac{\frac{w_2}{M_2}}{\frac{w_2}{M_2}+\frac{w_1}{M_1}}\)
- where w2 and M2 represent the mass in grams and molar mass of the solute respectively
- w1 and M1 represent the mass in grams and molar mass of the solvent respectively.
Here, we need to find out the mass of the non-volatile, non-electrolyte solute (w2).
Let the vapour pressure of the solvent (p0) be 100.
Since the vapour pressure will be 80% of 100, i.e., 80,
so, \(p_0-p_s=80\)
M2 = 8 g/mol, w1 = 114 g
The molar mass of octane, M1 (C8H18) = 12 x 8 + 1 x 18
Hence, M1 = 114 g
Using the formula for the relative lowering of vapour pressure, we can write:
\(\frac{80}{100}=\frac{\frac{w_2}{8}}{\frac{w_2}{8}+\frac{114}{114}}\)
\(⇒0.80=\frac{\frac{w_2}{8}}{\frac{w_2}{8}+1}\)
\(⇒\frac{w_2}{8}+1= \frac{w_2}{8}×0.80\)
\(⇒w_2+8=0.80w_2\)
\(⇒w_2-0.80w_2=8\)
\(⇒0.20w_2=8\)
\(⇒w_2=\frac{8}{0.20}=40g\)
Therefore, the mass of the non-volatile, non-electrolyte solute is 40g.
So, the correct option is (A): \(40\).
Approach Solution -3
Assuming the given solution is dilute.
By putting the values in the formula:
\(\dfrac{\Delta P}{ P^o_A} = \dfrac{n_B}{n_A} = \dfrac{w_B}{m_B} . \dfrac{m_A}{w_A}\)
\(\dfrac{20}{100} = \dfrac{8}{m_b}.\dfrac{114}{114}\)
\(m_B=\dfrac{8X100}{20}\)
\(=40\ gmol^{-1}\)
Hence, the correct answer is Option (A): \(40\)
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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