Question #185613

1. The molal freezing pointof water is 1.86°C/m(1.86K/m), which can be expressed as 1.86°C • kg H2O/mol solute) after some algebraic rearrangement. This means that a solution of 1 mol cane sugar (342g, over 3/4 lb) dissolvedin 1kgof water should freeze at -1.86°C.

Describe the freezing point drop suggestin number 1.

Expert's answer

"T \n_f\n\u200b\t\n = \\frac {1000\u00d7K \n_f\n\u200b\t\n \u00d7w} {342\u00d71000} =\n\u200b1.86K"

"T_f=0-1.86=-1.86K"

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