A company dissolves 'X' amount of CO_{2} at 2 | vedclass

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

Question

(D) According to Henry's Law: $P = K_{H} 	imes \chi_{CO_{2}}$Given:$P = 0.835 \ bar$$K_{H} = 1.67 \ kbar = 1670 \ bar$Since the amount of $CO_{2}$ is

Vedclass · Accepted Answer

$1222$

Vedclass · Answer

(D) According to Henry's Law: $P = K_{H} 	imes \chi_{CO_{2}}$Given:$P = 0.835 \ bar$$K_{H} = 1.67 \ kbar = 1670 \ bar$Since the amount of $CO_{2}$ is very small,we can approximate the mole fraction $\chi_{CO_{2}} \approx \frac{n_{CO_{2}}}{n_{H_{2}O}}$$n_{H_{2}O} = \frac{1000 \ g}{18 \ g \ mol^{-1}} = 55.55 \ mol$$0.835 = 1670 	imes \frac{n_{CO_{2}}}{55.55}$$n_{CO_{2}} = \frac{0.835 	imes 55.55}{1670} = 0.02777 \ mol$Mass of $CO_{2} = n_{CO_{2}} 	imes 	ext{Molar mass of } CO_{2} = 0.02777 	imes 44 = 1.222 \ g$$X = 1.222 	imes 10^{3} 	imes 10^{-3} \ g = 1222 	imes 10^{-3} \ g$Thus,$X = 1222$.

Similar Questions

Which of the following is an example of a solution in which the solute is a gas and the solvent is a solid?

How many millimoles of $CO_2$ gas will dissolve when $CO_2$ gas is passed in $900 \ mL$ water at $298 \ K$ temperature? $[$ The value of $K_H$ is $6.02 \times 10^{-4} \ bar$ and partial pressure of $CO_2$ gas is $2 \times 10^{-8} \ bar.]$

Identify a solution which contains a solid as the solute and a liquid as the solvent.

Rank the following solvents in decreasing order of their ability to dissolve the given solute (naphthalene).

If the $K_{H}$ values for $Ar_{(g)}$,$CO_{2_{(g)}}$,$HCHO_{(g)}$ and $CH_{4_{(g)}}$ respectively are $40.39$,$1.67$,$1.83 \times 10^{-5}$ and $0.413$,then identify the correct increasing order of their solubilities.

Vedclass Test Series

Exam Paper Generator

Online Exam Module