Asked by humphrey
A solution is prepared by dissolving 4.9g of sucrose in 17g water. Calculate the boiling point, freezing point and osmotic pressure of this solution at 25 degree Celsius.
Answers
Answered by
DrBob222
mols sucrose = grams/molar mass = 4.9/342 = ?
m = molality = mols sucrose/kg solvent = ?mols sucrose/0.017 kg = ?
dT = Kf*m. You have Kf for water and molality from above. Solve for delta T then normal freezing point - delta T = new freezing point. Kf is 1.86 for water.
dT = Kb*m
You have Kb for water (0.512), m from above, solve for delta T. Then normal boiling point for H2O + delta T = new boiling point.
osmotic pressure = MRT
You know T (remember to use kelvin), and R (universal gas constant). Convert molality (m) to molarity (M) and solve for osmotic pressure. For very dilute solutions m may be used interchangeably with M
Post your work if you get stuck.
m = molality = mols sucrose/kg solvent = ?mols sucrose/0.017 kg = ?
dT = Kf*m. You have Kf for water and molality from above. Solve for delta T then normal freezing point - delta T = new freezing point. Kf is 1.86 for water.
dT = Kb*m
You have Kb for water (0.512), m from above, solve for delta T. Then normal boiling point for H2O + delta T = new boiling point.
osmotic pressure = MRT
You know T (remember to use kelvin), and R (universal gas constant). Convert molality (m) to molarity (M) and solve for osmotic pressure. For very dilute solutions m may be used interchangeably with M
Post your work if you get stuck.
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