The Henry's law constant for the solubility o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) According to Henry's law,the partial pressure of a gas is given by $P_{N_2} = X_{N_2(air)} 	imes P_{total} = 0.8 	imes 5 \ atm = 4 \ atm$. Using

Vedclass · Accepted Answer

$4 	imes 10^{-4}$

Vedclass · Answer

(A) According to Henry's law,the partial pressure of a gas is given by $P_{N_2} = X_{N_2(air)} 	imes P_{total} = 0.8 	imes 5 \ atm = 4 \ atm$. Using Henry's law formula: $P_{N_2} = K_H 	imes X_{N_2(solute)}$. Substituting the values: $4 \ atm = (1.0 	imes 10^5 \ atm) 	imes X_{N_2(solute)}$. Therefore,$X_{N_2(solute)} = 4 	imes 10^{-5}$. Since the mole fraction $X_{N_2} = \frac{n_{N_2}}{n_{N_2} + n_{H_2O}} \approx \frac{n_{N_2}}{n_{H_2O}}$ (as $n_{N_2}$ is very small),we have $n_{N_2} = X_{N_2} 	imes n_{H_2O}$. $n_{N_2} = 4 	imes 10^{-5} 	imes 10 \ moles = 4 	imes 10^{-4} \ moles$.

The Henry's 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

Similar Questions

In the ground state,an element has $13$ electrons in the $M$ shell. The element is

Which of the following biomolecules is common to respiration-mediated breakdown of fats,carbohydrates and proteins?

Which one of the following structures between two adjacent cells is an effective transport pathway?

Which of the products is formed when acetone is reacted with barium hydroxide solution?

The main objective of the production/use of herbicide-resistant $GM$ crops is to

Vedclass Test Series

Exam Paper Generator

Online Exam Module