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What mass of NaCI ("molar mass" =58.5g m...

What mass of `NaCI ("molar mass" =58.5g mol^(-1))` be dissolved in `65g` of water to lower the freezing point by `7.5^(@)C`? The freezing point depression constant, `K_(f)`, for water is `1.86 K kg mol^(-1)`. Assume van't Hoff factor for `NaCI` is `1.87`.

Text Solution

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Here `M_(2) = 58.5 g mol^(-1), w_(1) = 65g, w_(2) = ?`
`DeltaT_(f) = 7.5^(@)C, i=1.87, K_(f) = 1.86 K kg mol^(-1)`
Using the formula, `DeltaT_(f) = iK_(f)m`
`DletaT_(f) = iK_(f) xx (w_(2) xx1000)/(M_(2)xxw_(1)) rArr 7.5 = 1.87 xx 1.86 xx (w_(2) xx 1000)/(58.5 xx 65)`
`:. w_(2) = (7.5 xx 58.5 xx 65)/(1.87 xx 1.86 xx 1000) = (28518.75)/(3478.2) = 8.188g`
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