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Calculate molality of 2.5 of ethanoic ac...

Calculate molality of 2.5 of ethanoic acid `(CH_(3)COOH)` in 75g of benzene.

Text Solution

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Molar mass of `C_2 H_4 O_2 = 12 xx 2 + 1 xx 4 + 16 xx 2 = 60"g mol"^(-1)`
Moles mass of `C_(2) H_(4) O_(2) = (2.5"g")/(60"g mol"^(-1) ) = 0.0417" mol"`
Mass of benzene in kg `= 75 g// 1000" g kg"^(-1) = 75 xx 10^(-3)" kg"`
Molality of `C_(2) H_(4) O_(2) = ("moles of" C_(2) H_(4) O_(2) )/("kg of benzene")`
`= (0.0417" mol" xx 1000" g kg"^(-1) )/( 75" g")`
`= 0.556" mol kg"^(-1)`
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