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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. |
Answer» Correct Answer - 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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