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Calculate the amount of CaCl_2[molar mass = 111 g "mol"^(-1) ] which must be added to 500 g of water to lower the freezing point by 2 K, assuming CaCl_2is completely dissociated. |
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Answer» Solution :APPLY the relation (without taking dissociation into CONSIDERATION) ` Delta T_f = (K_f xx 1000 xx w_2)/(M_2 xx w_1)` `2K = (1.86 K KG "MOL"^(-1) xx 1000 g kg^(-1) xx w_2)/(111 g "mol"^(-1) xx 500 g) ` ` w_2 = (2 xx 111 xx 500)/(1.86 xx 1000) = 59.68 g` One molecule of `CaCl_2` dissociates into one `Ca^(2+)` ion and two `Cl^-` ions. `CaCl_2 to Ca^(2+) + 2Cl^(-)` PARTICLE dissociates into three particles. Therefore, the amount of `CaCl_2`needed = `(59.68)/(3) = 19.89 g` |
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