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In 200 g of water 0.01 mole of NaCI and 0.02 mole of sucrose are dissolved. Assuming solution to be ideal, the depression in freezing point of water (in oC) will be ________. (final answer should be rounded off to two decimal places) [Given : Kf(H2O) = 1.86 K kg mol–1]
Correct answer is '0.37'. Can you explain this answer?
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In 200 g of water 0.01 mole of NaCI and 0.02 mole of sucrose are disso...
Solution:
Given:
Mass of water (w) = 200 g
Moles of NaCl (n1) = 0.01 mole
Moles of sucrose (n2) = 0.02 mole
Kf(H2O) = 1.86 K kg mol-1
The depression in freezing point of water (ΔTf) is given by the formula:
ΔTf = Kf × w2 × (n1 + n2)
where w2 is the molecular weight of solvent in kg mol-1.
Calculation:
Molecular weight of water (w2) = 18.015 g mol-1 / 1000 = 0.018015 kg mol-1
Molecular weight of NaCl = 58.44 g mol-1
Molecular weight of sucrose = 342.3 g mol-1
Total moles of solute (n1 + n2) = 0.01 + 0.02 = 0.03 mol
Substituting the values in the formula, we get:
ΔTf = 1.86 × 0.2 × 0.03 / 0.018015 = 0.37 K
Therefore, the depression in freezing point of water is 0.37 oC.
Hence, the answer is '0.37'.
Given:
Mass of water (w) = 200 g
Moles of NaCl (n1) = 0.01 mole
Moles of sucrose (n2) = 0.02 mole
Kf(H2O) = 1.86 K kg mol-1
The depression in freezing point of water (ΔTf) is given by the formula:
ΔTf = Kf × w2 × (n1 + n2)
where w2 is the molecular weight of solvent in kg mol-1.
Calculation:
Molecular weight of water (w2) = 18.015 g mol-1 / 1000 = 0.018015 kg mol-1
Molecular weight of NaCl = 58.44 g mol-1
Molecular weight of sucrose = 342.3 g mol-1
Total moles of solute (n1 + n2) = 0.01 + 0.02 = 0.03 mol
Substituting the values in the formula, we get:
ΔTf = 1.86 × 0.2 × 0.03 / 0.018015 = 0.37 K
Therefore, the depression in freezing point of water is 0.37 oC.
Hence, the answer is '0.37'.
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In 200 g of water 0.01 mole of NaCI and 0.02 mole of sucrose are dissolved. Assuming solution to be ideal, the depression in freezing point of water (in oC) will be ________. (final answer should be rounded off to two decimal places) [Given : Kf(H2O) = 1.86 K kg mol–1]Correct answer is '0.37'. Can you explain this answer?
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