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

When electric current is passed through acidified water for $1930 \ s$,$1120 \ mL$ of $H_2$ gas is collected (at $STP$) at the cathode. What is the current passed in amperes?

$\int(x+1)^2 e^x d x$ is equal to

In which of the following molecules is the $N$ atom not $sp^2$ hybridized?

The number of points in $(-\infty, \infty)$ for which $x^2 - x \sin x - \cos x = 0$ is

The angle between the asymptotes of the hyperbola $x^2-3y^2=3$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module