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How will you prepare 1.0 kg of an aqueou...

How will you prepare 1.0 kg of an aqueous solution of acetone `(CH_(3)COOH_(3))` in water in which mole fraction of acitone is 0.19.

Text Solution

Verified by Experts

The correct Answer is:
Mass of acetone = 431 g; Mass of water=569 g
`"By definition :" " "x_(XH_(3)XOOH_(3))=(eta_(CH_(3)COOH_(3)))/((eta_(CH_(3)COOH_(3)))+eta_((H_(2)O)))`
`"Molar mass of "CH_(3)COOH_(3)=58" g mol"^(-1), "Molar mass of "H_(2)O=18" g mol"^(-1)`
`eta_((CH_(3)COOH_(3)))=("Mass of "CH_(3)COOH_(3))/("Molar mass of"CH_(3)COCH_(3))=((xg))/((58"g mol"^(-1)))=x/58mol`
`eta_((H_(2)O))=("Mass of water")/("Molar mass of water")=((1000-x))/((18"g mol"^(-1)))=((1000-x))/18mol`
`x_((CH_(3)COCH_(3)))=0.19("Given")`
`0.19=((x//58"mol"))/((x//58"mol")+(1000-x)//18"mol")=(x//58)/(x//58+(1000-x)//18)`
`0.19=((x//58)xx(58xx18))/(18x+58(1000-x))=(18x)/(18x+58000-58x)`
0.19 (58000- 40x) = 18 x
11020=7.6xx=18x or 25.6x=11020
`or" " x=(11020)/(25.6)=431 g`
`therefore" Mass of acetone to be added"=431 g`
Mass of water to be added =(1000-431)=569 g.
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