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An anti-freeze solution is prepared from 222.6 g of ethylene glycol C_(2) H_(4) (OH)_(2) and 200 g of water . Calculate themolality of the solution. If the density of this solution be 1.072 g mL^(-1) , what will be the molarity of the solution ? |
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Answer» Solution :Mg of `C_(2) H_(4) (OH)_(2) = 62 g mol^(-1)` Molality = `(n_(B))/(W_(A))xx 1000 = (W_(B))/(M_(B) xx W_(B)) xx 1000 = (222.6 xx 1000)/(62 xx 200) = 17.95 m ` DENSITY= `("Mass")/("Volume")` So , Volume = `("Mass")/("Density") = (422.6)/(1.072) = 394.22` ml `M = (n_(B))/(V) xx 1000` `= (222.6)/(394.22 xx 62) xx 1000 = 9.11` M |
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