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What mass of ethlene ethyelne glycol `("molar mass"=62 g mol^(-1))` must be dissolved in 5.5kg of water to lower the freezing point of from `0^(@)C to-10^(@)C` ? `(K_(f))` for water=1.86 K kg `mol^(-1)`). |
Answer» Here `M_(2) = 62.0" g " mol^(-1) , w_(1) = 5.50 kg` ` = 5500 g, w_(2)` = ? `K_(f) = 1.86" K " kg mol^(-1)` `Delta T _(f) = 0-(-10.0) = 10^(@) C or 10 K` Using the formual ` Delta T_(f) = K_(f) m = K_(f) m = K_(f) xx (w_(2) xx 1000)/(M_(2) xx w_(1)) ` ` 10 = (1.86 xx w_(2) xx 1000)/(62.0 xx 5500)` ` :. " "w_(2) = (10 xx 62.0 xx 5500)/(1.86 xx 1000) = (62 xx 55)/(1.86) = 1833.33" g " = 1.833 kg`. |
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