Answer to Question #251980 in General Chemistry for Larah

Question #251980
A solution consists of 10.3 g of the nonelectrolyte glucose,C6H12O6 dissolved in 250.0 g of water. What is the freezing point depression of the solution?
1
Expert's answer
2021-10-16T02:01:44-0400

The freezing point depression

"\u0394T_f =k_fm \\\\\n\nk_f = 1.86 \\;\u00b0C\/m"

Molality

"m = \\frac{Number \\;of \\;moles\\; of\\; glucose}{Weight \\;of \\;solvent\\; in\\; kg}"

M(glucose) = 180 g/mol

Molality

"n(glucose) =\\frac{10.3}{180} = 0.05722 \\\\\n\nm= \\frac{0.05722}{250 \\times 10^{-3}} = 0.2288 \\;m \\\\\n\n\u0394T_f =1.86 \\times 0.2288 = 0.425 \\;\u00b0C"


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