Calculate K_H when 2 mmol of O_2 dissolve in | vedclass

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

Question

(C) Henry's Law is given by $P = K_H 	imes \chi$,where $\chi$ is the mole fraction of the gas in the solution. First,calculate the moles of water: $n

Vedclass · Accepted Answer

$1 	imes 10^{-4} \ bar$

Vedclass · Answer

(C) Henry's Law is given by $P = K_H 	imes \chi$,where $\chi$ is the mole fraction of the gas in the solution. First,calculate the moles of water: $n_{H_2O} = \frac{540 \ g}{18 \ g/mol} = 30 \ mol$. Since $2 \ mmol = 2 	imes 10^{-3} \ mol$ is very small,the total moles $\approx 30 \ mol$. Mole fraction $\chi_{O_2} = \frac{n_{O_2}}{n_{H_2O}} = \frac{2 	imes 10^{-3}}{30} = \frac{2}{3} 	imes 10^{-4} \approx 0.667 	imes 10^{-4}$. Using $K_H = \frac{P_{O_2}}{\chi_{O_2}} = \frac{2 	imes 10^{-8}}{0.667 	imes 10^{-4}} = 3 	imes 10^{-4} \ bar$.

Calculate $K_H$ when $2 \ mmol$ of $O_2$ dissolve in $540 \ mL$ of water at $27 \ ^oC$ temperature. Given $P_{O_2} = 2 \times 10^{-8} \ bar$.

Similar Questions

The variation of solubility with temperature $T$ for a gas in a liquid is shown by the following graphs. The correct representation is:

The values of the Henry's law constant of $Ar$,$CO_{2}$,$CH_{4}$ and $O_{2}$ in water at $25^{\circ} C$ are $40.30$,$1.67$,$0.41$ and $34.86 \ kbar$,respectively. The order of their solubility in water at the same temperature and pressure is

Which of the following laws represents the quantitative relationship between the solubility of gas in liquid and its pressure?

Choose the correct statement.

At $298 \ K$,the mole percentage of $N_{2(g)}$ in air is $80\%$. Water is in equilibrium with air at a pressure of $10 \ atm$. What is the mole fraction of $N_{2(g)}$ in water at $298 \ K$? ($K_H$ for $N_2$ is $6.5 \times 10^7 \ mm \ Hg$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module