Question:
The Henry?? law constant for the solubility of $N_2$ gas in water at $298\, K$ is $1.0 \times 10^5\, atm$. The mole fraction of $N_2$ in air is $0.8$. The number of moles of $N_2$ from air dissolved in $10$ moles of water at $298\, K$ and $5 \,atm$ pressure is
The Henry?? law constant for the solubility of $N_2$ gas in water at $298\, K$ is $1.0 \times 10^5\, atm$. The mole fraction of $N_2$ in air is $0.8$. The number of moles of $N_2$ from air dissolved in $10$ moles of water at $298\, K$ and $5 \,atm$ pressure is
Updated On: Aug 1, 2022
- $4.0 \times 10^{-4}$
- $4.0 \times 10^{-5}$
- $5.0 \times 10^{-4}$
- $4.0 \times 10^{-6}$
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The Correct Option is A
Solution and Explanation
According to Henry?? law,
$x_{N_2} \times K_H = p_{N_2}$ ($p_{N_2} =$ Partial pressure of $N_2$)
Given, total pressure $= 5\, atm$
mole fraction of $N_2 = 0.8$
$ \therefore$ Partial pressure of $N_{2} = 0.8 \times 5 = 4$
$\Rightarrow x_{N_2} \times 1 \times 10^{5} = 4$
$\Rightarrow x_{N_2} = 4 \times 10^{-5}$
no. of moles of $H_{2}O$, $n_{H_2O} = 10$
no. of nmoles of $N_{2}$, $n_{N_2 }= ?$
$\frac{n_{N_2}}{n_{N_2}+ n_{H_2O}} = x_{N_2}=4 \times 10^{-5}$
$\Rightarrow \frac{n_{N_2}}{10 +n_{N_2}} = 4\times10^{-5}$
$\Rightarrow n_{N_2} = 4 \times 10^{-4} \quad\left[\because n_{N_2} <<< 10\right]$
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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