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Two solutions A and B are prepared by dissolving 1 g of non-volatile solutes X and Y respectively in 1 kg of water. The ratio of depression in freezing points for A and B is found to be 1 : 4. The ratio of molar masses of X and Y is:
  • a)
    1 : 4
  • b)
    1 : 0.25
  • c)
    1 : 0.20
  • d)
    1 : 5
Correct answer is option 'B'. Can you explain this answer?
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Given Information:

- Mass of solute X = 1 g
- Mass of solute Y = 1 g
- Mass of water = 1 kg
- Ratio of depression in freezing points for A and B = 1:4

Calculations:

Let Molar mass of X = mX g/mol
Let Molar mass of Y = mY g/mol

Moles of X = 1/mX
Moles of Y = 1/mY

Given that the ratio of depression in freezing points for A and B is 1:4, we have:

ΔTf, A / ΔTf, B = 1 / 4

Using the formula for depression in freezing point:

ΔTf = Kf * molality

Where Kf is the cryoscopic constant of water.

Since the molality is given by:

molality = (moles of solute) / (mass of solvent in kg)

We can write the equation for the ratio of depression in freezing points as:

(mX / 1) / (mY / 1) = 1 / 4

Simplifying, we get:

mX / mY = 1 / 4

Therefore, the ratio of molar masses of X and Y is 1:4, which can be written as 1:0.25 in decimals. Hence, the correct answer is option 'b) 1:0.25'.
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Two solutions A and B are prepared by dissolving 1 g of non-volatile solutes X and Y respectively in 1 kg of water. The ratio of depression in freezing points for A and B is found to be 1 : 4. The ratio of molar masses of X and Y is:a)1 : 4b)1 : 0.25c)1 : 0.20d)1 : 5Correct answer is option 'B'. Can you explain this answer?
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