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An antifreeze solution is prepared from 222.6 g of ethylene glycol [C_2H_4 (OH)_2] and 200 g of water. Calculate the molality 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 : APPLYING the following relation and SUBSTITUTING the VALUES, we get `m = (W_B)/(M_B) XX (1000)/(W_A)= (222.6)/(62) xx 1000/200 = 2226/124 = 17.95`mol/kg Molality (m) and molarity (M), are RELATED as follows : `m = (M)/(d - M xx M_B xx 10^(-3))` ` 17.95 = (M)/(1.072 - M xx 62 xx 10^(-3))` `17.95 xx 1.072 - 17.95 xx 62 xx 10^(-3) M = M` `M (1+ 17.95 xx 62 xx 10^(-3)) = 19.24 " or "M = (19.24)/(2.11) = 9.12` mol/litre Molarity of the solution = 9.12 M. |
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