At T (K) the molarity of CO_2 (in mol L^{-1 | vedclass

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

Question

(B) According to Henry's law,$p = K_H \cdot \chi$,where $\chi$ is the mole fraction of $CO_2$. Given $p = 3.4 \ bar$ and $K_H = 1.7 	imes 10^3 \ bar$

Vedclass · Accepted Answer

$1.11 	imes 10^{-1}$

Vedclass · Answer

(B) According to Henry's law,$p = K_H \cdot \chi$,where $\chi$ is the mole fraction of $CO_2$. Given $p = 3.4 \ bar$ and $K_H = 1.7 	imes 10^3 \ bar$. $\chi = \frac{p}{K_H} = \frac{3.4}{1.7 	imes 10^3} = 2 	imes 10^{-3}$. Since the solution is dilute,$\chi = \frac{n_{CO_2}}{n_{H_2O}}$. $n_{H_2O} = \frac{200 \ g}{18 \ g \ mol^{-1}} = 11.11 \ mol$. $n_{CO_2} = \chi 	imes n_{H_2O} = 2 	imes 10^{-3} 	imes 11.11 = 2.222 	imes 10^{-2} \ mol$. Molarity $= \frac{n_{CO_2}}{	ext{Volume in } L} = \frac{2.222 	imes 10^{-2} \ mol}{0.2 \ L} = 0.1111 \ mol \ L^{-1} = 1.11 	imes 10^{-1} \ mol \ L^{-1}$.

Gas	$K_{H} \ (kbar)$
$\alpha$	$50$
$\beta$	$2$
$\gamma$	$2 \times 10^{-5}$
$\delta$	$0.5$

At $T \ (K)$ the molarity of $CO_2$ (in $mol \ L^{-1}$) in $200 \ mL$ of soda water packed under a pressure of $3.4 \ bar$ is ($K_H$ of $CO_2$ in water is $1.7 \times 10^3 \ bar$ at $T \ (K)$).

Similar Questions

Maximum amount of a solid solute that can be dissolved in a specified amount of a given liquid solvent does not depend upon $ . . . . . . $ .
$(i)$ Temperature $(ii)$ Nature of solute $(iii)$ Pressure $(iv)$ Nature of Solvent

The statement “If $0.003 \ mol$ of a gas are dissolved in $900 \ g$ of water under a pressure of $1 \ atm$,$0.006 \ mol$ will be dissolved under a pressure of $2 \ atm$”,illustrates:

$1 \ g$ of silver gets distributed between $10 \ cm^{3}$ of molten zinc and $100 \ cm^{3}$ of molten lead at $800^{\circ} C$. The percentage of silver still left in the lead layer is approximately

Explain the equilibrium in soda water and explain it using Henry's law.

Vedclass Test Series

Exam Paper Generator

Online Exam Module