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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.

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Molar mass of `CH_(3) COOH = 2 xx 12 +4xx 1+2xx 16=60 g mol^(-1)`
Mass of benzene = 75 g
Molality of `CH_(3)COOH = ("Moles of " CH_(3)COOH)/("Mass of benzene (in g)") xx 1000`
` = (0.0417 mol)/(75) xx 1000`
` = 0.556 m`
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