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

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Text Solution

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The correct Answer is:
`1.854g`
`K_(H)=1.67xx10^(8)Pa`
`p_(CO_(2))=2.5atm=2.5xx101325Pa`
Applying Henry's law :
`CHMi_(CO_(2))=(p_(CO_(2)))/(K_(H))=(2.5xx101325Pa)/(1.67xx10^(8)Pa)=1.517xx10^(-3)`
`CHMi_(CO_(2))=(n_(CO_(2)))/(n_(CO_(2))+n_(H_(2)O))~~(n_(CO_(2)))/(n_(H_(2)O))=1.517xx10^(-3)`...... (i)
For `500mL` of soda water, water present
`=500mL=500g`
`=(500)/(18)=27.78 mol of H_(2)O`
`:. n_(H_(2)O)=27.78`
Substitute the value of `n_(H_(2)O)` in `Eq.(i),` we get
`(n_(CO_(2)))/(27.78)=1.517xx10^(-3)`
`:. n_(CO_(2))=1.517xx10^(-3)xx27.78mol`
`=42.14xx10^(-3)mol`
`Mw` of `CO_(2)=12+2xx16=44g mol^(-1)`
`:. ` Weight of `CO_(2)=mol xx Mw`
`=42.14xx10^(-3)xx44g=1.854g`
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