Home
Class 12
CHEMISTRY
A solution of urea (mol. Mass 60 g mol^(...

A solution of urea (mol. Mass `60 g mol^(-1)`) boils of `100.18^(@)C` at one one atmospheric pressure. If `k_(f)` and `K_(b)` for water are `1.86` and `0.512 K kg mol^(-1)` respectively, the above solution will freeze at:

A

`-6.54^@C`

B

`6.54^@C`

C

`0.654^@C`

D

`-0.654^@C`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the freezing point of the urea solution using the given data about its boiling point elevation and the freezing point depression constants for water. ### Step-by-step Solution: 1. **Identify the Given Data:** - Molecular mass of urea = 60 g/mol - Boiling point of the solution = 100.18 °C - Boiling point of pure water = 100 °C - \( K_f \) (freezing point depression constant for water) = 1.86 °C kg/mol - \( K_b \) (boiling point elevation constant for water) = 0.512 °C kg/mol 2. **Calculate the Boiling Point Elevation (\( \Delta T_b \)):** \[ \Delta T_b = \text{Boiling point of solution} - \text{Boiling point of pure solvent} = 100.18 °C - 100 °C = 0.18 °C \] 3. **Use the Relationship Between Freezing Point Depression and Boiling Point Elevation:** The relationship between the freezing point depression (\( \Delta T_f \)) and boiling point elevation (\( \Delta T_b \)) can be expressed as: \[ \frac{\Delta T_f}{\Delta T_b} = \frac{K_f}{K_b} \] 4. **Substitute the Known Values:** \[ \frac{\Delta T_f}{0.18} = \frac{1.86}{0.512} \] 5. **Calculate \( \Delta T_f \):** \[ \Delta T_f = 0.18 \times \frac{1.86}{0.512} \] \[ \Delta T_f = 0.18 \times 3.6328125 \approx 0.654 °C \] 6. **Determine the Freezing Point of the Solution:** The freezing point depression is given by: \[ \Delta T_f = T_f (\text{pure solvent}) - T_f (\text{solution}) \] Since the freezing point of pure water is 0 °C, we can set up the equation: \[ 0.654 °C = 0 °C - T_f (\text{solution}) \] Rearranging gives: \[ T_f (\text{solution}) = 0 °C - 0.654 °C = -0.654 °C \] 7. **Final Result:** The freezing point of the urea solution is approximately: \[ T_f (\text{solution}) \approx -0.65 °C \] ### Conclusion: The freezing point of the urea solution is approximately -0.65 °C.
Promotional Banner

Similar Questions

Explore conceptually related problems

A solution of urea boils at 100.18^(@)C at the atmospheric pressure. If K_(f) and K_(b) for water are 1.86 and 0.512K kg mol^(-1) respectively, the above solution will freeze at,

A solution of urea in water has boiling point of 100.15^(@)C . Calculate the freezing point of the same solution if K_(f) and K_(b) for water are 1.87 K kg mol^(-1) and 0.52 K kg mol^(-1) , respectively.

An aqueous solution freezes on 0.36^@C K_f and K_b for water are 1.8 and 0.52 k kg mol^-1 respectively then value of boiling point of solution as 1 atm pressure is

What do you understand by the term that K_(f) for water is 1.86 K kg mol^(-1) ?

A solution of urea in water has a boiling point of 100*18^@C . Calculate the freezing point of the solution.(K for water is 1*86 " K kg " mol^(-1) and K_b " for water is " 0*512 " K kg " mol^(-l) ).

A solution containing 0.11kg of barium nitrate in 0.1kg of water boils at 100.46^(@)C . Calculate the degree of ionization of the salt. K_(b) (water) = 0.52 K kg mol^(-1) .

Pure water boils at 99.725^(@)C at Shimla. If K_(b) for water is 0.51 K mol^(-1) kg the boiling point of 0.69 molal urea solution will be:

For a 5% solution of urea (Molar mass - 60 g/mol), calculate the osmotic pressure at 300 K. [R = 0.0821 L atm K^(-1) mol^(-1)]

Complete the following statements by selecting the correct alternative from the choices given : An aqueous solution of urea freezes at - 0.186^(@)C, K_(f) for water = 1.86 K kg. mol^(-1),K_(b) for water = 0.512 "K kg mol"^(-1) . The boiling point of urea solution will be :

1.00 g of a non-electrolyte solute (molar mass 250g mol^(–1) ) was dissolved in 51.2 g of benzene. If the freezing point depression constant K_(f) of benzene is 5.12 K kg mol^(–1) , the freezing point of benzene will be lowered by:-