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Chemistry question about Freezing Points?
You use 6.200 g of solvent (FP = 73.3°C; Kf = 4.5°C/m) and 0.310 g of solute (molecular mass = 432.11 g/mol).
(a) How many moles of solute have you used?
(b) What is the colligative molality of the solution formed?
(c) What is the predicted freezing point depression ΔTf?
(d) What is the predicted freezing point of the solution formed?
to clarify I really need help with parts C & D: here are the answers to part A & B and how to find them:
(a) moles solute is .310g/431.11g/mol=7.190e-4 moles
(b) colligative molality is 7.190e-4/.0062=.116 m
1 Answer
- BobbyLv 78 years agoFavorite Answer
moles of solute = .310 / 432.11 = 0.00071741 moles
these are dissolved in 6.200 g of solvent so we have molality = 0.00071741 x 1000 / 6.200 moles / kg of solvent = 0.1157 m
delta t = m x Kf x i
= 0.1157 x 4.5 x 1 = .52 o C
new freezing point = 73.3 - .520 = 72.8 oC
