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One molal aqueous solution of urea freezes at -1.86° C. Aqueous solution of CnH2nOn 36% by mass of water freezes at -3.72° C. What is the value of n?
Correct answer is '6'. Can you explain this answer?
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One molal aqueous solution of urea freezes at -1.86° C. Aqueous s...
(6) Molality of urea solution = 1 mol kg-1
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One molal aqueous solution of urea freezes at -1.86° C. Aqueous s...
Explanation:
Colligative properties:
Colligative properties are the properties of a solution that depend only on the number of solute particles in it, not on the nature of the solute or the solvent. The four colligative properties are:
- Lowering of vapor pressure
- Elevation of boiling point
- Depression of freezing point
- Osmotic pressure
Depression of freezing point:
The depression of freezing point is the difference between the freezing point of the pure solvent and the freezing point of the solution. It is given by the formula:
ΔTf = Kf × molality
where, ΔTf is the depression of freezing point, Kf is the freezing point depression constant of the solvent, and molality is the molality of the solution.
Calculations:
Given:
Molality of urea solution (m1) = 1 molal
Freezing point depression of urea solution (ΔTf1) = -1.86 C
Molality of CnH2nOn solution (m2) = 36% by mass of water
Freezing point depression of CnH2nOn solution (ΔTf2) = -3.72 C
We need to find the value of n.
Molality of urea solution (m1) = 1 molal
Freezing point depression of urea solution (ΔTf1) = -1.86 C
Molality of CnH2nOn solution (m2) = 36% by mass of water
Freezing point depression of CnH2nOn solution (ΔTf2) = -3.72 C
We need to find the value of n.
For urea solution:
ΔTf1 = Kf × m1
-1.86 = Kf × 1
Kf = -1.86
ΔTf1 = Kf × m1
-1.86 = Kf × 1
Kf = -1.86
For CnH2nOn solution:
Molar mass of water (M1) = 18 g/mol
Molar mass of CnH2nOn (M2) = 12n + 1(2n) + 16n = 28n + 12 g/mol
Mass of water (m1) = 36 g
Mass of CnH2nOn (m2) = 64 g
Number of moles of water (n1) = m1/M1 = 2
Number of moles of CnH2nOn (n2) = m2/M2 = 2.29n
Molality of CnH2nOn solution (m2) = n2/(n1+m1/1000) = 1.04 molal
ΔTf2 = Kf × m2
-3.72 = -1.86 × 1.04
n = 6
Molar mass of water (M1) = 18 g/mol
Molar mass of CnH2nOn (M2) = 12n + 1(2n) + 16n = 28n + 12 g/mol
Mass of water (m1) = 36 g
Mass of CnH2nOn (m2) = 64 g
Number of moles of water (n1) = m1/M1 = 2
Number of moles of CnH2nOn (n2) = m2/M2 = 2.29n
Molality of CnH2nOn solution (m2) = n2/(n1+m1/1000) = 1.04 molal
ΔTf2 = Kf × m2
-3.72 = -1.86 × 1.04
n = 6
Answer:
The value of n is 6.
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One molal aqueous solution of urea freezes at -1.86° C. Aqueous solution of CnH2nOn 36% by mass of water freezes at -3.72° C. What is the value of n?Correct answer is '6'. Can you explain this answer?
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