1.

Henry's law constant for CO_(2) in water is 1.67xx10^(8) Pa at 298 K. Calculate the quantity of CO_(2) in 500 mL of soda water when packed under 2.5 atm CO_(2) pressure at 298 K.

Answer»

SOLUTION :`K_(H)=1.67xx10^(8)"Pa,"p_(CO_(2))="2.5 atm"=2.5xx101325Pa`
`"Applying Henry's law, "p_(CO_(2))=K_(H)xx x_(CO_(2))`
`therefore""x_(CO_(2))=(p_(CO_(2)))/(K_(H))=(2.5xx101325Pa)/(1.67xx10^(8)Pa)=1.517xx10^(-3),"i.e.,"(n_(CO_(2)))/(n_(H_(2)O)+n_(CO_(2)))~=(n_(CO_(2)))/(n_(H_(2)O))=1.517xx10^(-3)`
`"For 500 mL of SODA water, water present "~="500 mL = 500 G "=(500)/(18)="27.78 moles"`
i.e.,` n_(H_(2)O)="27.78 moles"`
`therefore (n_(CO_(2)))/(27.78)=1.517xx10^(-3)or n_(CO_(2))=42.14xx10^(-3)" mole = 42.14 m mol "=42.14xx10^(-3)xx44g=1.854g.`


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