**Please Help Me**A solution is made by dissolving 15.5 grams of glucose (C6H12O6) in 245 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem.
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use the formula \(\Delta T=i*m*K_f\)
where \(\Delta T\) is the change in temp from the pure solvent's f.p.
i is the van't hoff constant, 1 in the case for sugar
m is the molality of the solution
\\(K_f\) is the f.p. depression constant
wait don't we got to find the morality first @aaronq
then use the formula
Can you setup how to find the moles for me
divide the mass by the molar mass of glucose (which is 180 g/mol)
Ok I have 0.15996
what is that?
Delta T @aaronq
So that is the freezing point
idk, it mean it sounds right. you found moles, then molality, then multiplied by the constant?
if so, then it should be right.