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60g of urea is dissolved in 1100g soluti...

60g of urea is dissolved in 1100g solution. To keep `DeltaT//K_(f)" as "1" mol"//kg`, water separated in the form of ice is

A

40 g

B

60 g

C

100 g

D

200 g

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To solve the problem, we need to determine how much water is separated in the form of ice when 60 g of urea is dissolved in a solution weighing 1100 g, given that the depression in freezing point (ΔTf) is equal to 1 mol/kg. ### Step-by-Step Solution: 1. **Understand the given data:** - Mass of urea (solute) = 60 g - Mass of solution = 1100 g - Depression in freezing point (ΔTf) = 1 K - Kf (cryoscopic constant for water) = 1 mol/kg (as per the problem) 2. **Calculate the molality (m):** - The formula for depression in freezing point is given by: \[ \Delta T_f = K_f \times m \] - Rearranging gives us: \[ m = \frac{\Delta T_f}{K_f} = \frac{1}{1} = 1 \text{ mol/kg} \] 3. **Calculate the number of moles of urea:** - The molecular weight of urea (NH2CONH2) is approximately 60 g/mol. - Number of moles of urea: \[ \text{Number of moles} = \frac{\text{mass of urea}}{\text{molar mass}} = \frac{60 \text{ g}}{60 \text{ g/mol}} = 1 \text{ mol} \] 4. **Determine the weight of the solvent (water):** - The total mass of the solution is 1100 g, which includes both the solute (urea) and the solvent (water). - Weight of solvent (water): \[ \text{Weight of solvent} = \text{Weight of solution} - \text{Weight of solute} = 1100 \text{ g} - 60 \text{ g} = 1040 \text{ g} \] 5. **Calculate the amount of water that remains after freezing:** - We know that the molality is 1 mol/kg, which means there is 1 kg (1000 g) of water for every mole of solute. - Since we have 1 mole of urea, this means we need 1000 g of water to achieve the desired molality. - Therefore, the amount of water that remains in liquid form is 1000 g. 6. **Calculate the amount of water that is separated as ice:** - The total weight of the solvent is 1040 g, and the weight of water that remains is 1000 g. - Thus, the weight of water that has separated as ice is: \[ \text{Weight of ice} = \text{Total weight of solvent} - \text{Weight of water remaining} = 1040 \text{ g} - 1000 \text{ g} = 40 \text{ g} \] ### Final Answer: The amount of water separated in the form of ice is **40 g**.
To solve the problem, we need to determine how much water is separated in the form of ice when 60 g of urea is dissolved in a solution weighing 1100 g, given that the depression in freezing point (ΔTf) is equal to 1 mol/kg. ### Step-by-Step Solution: 1. **Understand the given data:** - Mass of urea (solute) = 60 g - Mass of solution = 1100 g - Depression in freezing point (ΔTf) = 1 K ...
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