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Calculate molality of 2.5 of ethanoic acid `(CH_(3)COOH)` in 75g of benzene. |
Answer» Moles of `C_(2)H_(4)O_(2):12xx2+1xx4+16xx2=60"g mol"^(-1)` Moles of `C_(2)H_(4)O_(2)=(2.5" g")/(60" g mol"^(-1))=0.0417" mol"` Mass of benzene is kg `=75" g"//1000" g kg"^(-1)=75xx10^(-3)" kg"` Molality of `C_(2)H_(4)O_(2)=("Moles of "C_(2)H_(4)O_(2))/("kg of benzene")=(0.0417xx" mol "xx1000" g kg"^(-1))/(75"g")` `=0.556" mol kg"^(-1)` |
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