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Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g "mol"^(-1) ) |
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Answer» Solution :APPLY the FORMULA ` DELTA T_f = (K_f xx w_2 xx 1000)/(M_2 xx w_1)` substituting the VALUES in the above equation , we have ` Delta T_f = (1.86 K kg "mol"^(-1) xx 60g xx 1000 g kg^(-1))/(180 g "mol"^(-1) xx 250 g)` `Delta T_f= (1.86 xx 60 xx 1000)/(180 xx 250) = 2.48 K` Freezing point of the solution ` = 273.15 K - 2.48 K = 270.67 K` |
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