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A 5% solution (w/W) of cane sugar (molar mass = 342 g mol-1) has freezing point of 271 K. What will be the freezing point of a 5% glucose (molar mass = 18 g mol-1) in water if freezing point of pure water is 273.15 K?
  • a)
    273.07 K
  • b)
    269.07 K
  • c)
    273.15 K
  • d)
    260.09 K
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A 5% solution (w/W) of cane sugar (molar mass = 342 g mol-1)has freezi...

For cane sugar solution, 2.15 K = 
For glucose solution,


Freezing point of glucose solution = 273.15 - 4.085 = 269.07 K
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Most Upvoted Answer
A 5% solution (w/W) of cane sugar (molar mass = 342 g mol-1)has freezi...
Given:
- Concentration of cane sugar solution = 5% (w/w)
- Molar mass of cane sugar (sucrose) = 342 g/mol
- Freezing point of cane sugar solution = 271 K
- Concentration of glucose solution = 5% (w/w)
- Molar mass of glucose = 18 g/mol
- Freezing point of pure water = 273.15 K

To find:
Freezing point of the glucose solution

Solution:

1. Calculation of Molality (m):
Molality (m) is defined as the number of moles of solute per kilogram of solvent.
We need to calculate the molality of the cane sugar solution.

- Mass of cane sugar (sucrose) = 5% of the total mass of the solution
= 5/100 * mass of the solution

- Mass of water = Total mass of the solution - Mass of cane sugar
= Total mass of the solution - (5/100 * mass of the solution)

- Moles of cane sugar (sucrose) = Mass of cane sugar / Molar mass of cane sugar
= (5/100 * mass of the solution) / 342

- Moles of water = Mass of water / Molar mass of water
= (Total mass of the solution - (5/100 * mass of the solution)) / 18

- Molality (m) = Moles of cane sugar / Mass of water in kg
= [(5/100 * mass of the solution) / 342] / [(Total mass of the solution - (5/100 * mass of the solution)) / 18 * 1000]

2. Calculation of Freezing Point Depression (∆Tf):
Freezing point depression is the difference between the freezing point of a pure solvent and the freezing point of the solution.

∆Tf = Kf * m

where Kf is the cryoscopic constant.

3. Calculation of Freezing Point of the Glucose Solution:
The freezing point of the glucose solution can be calculated using the formula:

Freezing point of the solution = Freezing point of pure water - ∆Tf

Substituting the values:
- Freezing point of pure water = 273.15 K
- Cryoscopic constant (Kf) for water = 1.86 K kg/mol (given)

∆Tf = Kf * m

Freezing point of the solution = 273.15 K - ∆Tf

Calculating the values:
- From step 1, we have calculated the value of molality (m).
- Cryoscopic constant (Kf) for water = 1.86 K kg/mol

∆Tf = 1.86 * m

Freezing point of the solution = 273.15 K - 1.86 * m

Substituting the value of molality:
- From step 1, we have calculated the value of molality (m).

Let's consider the value of m as x.

Freezing point of the solution = 273.15 K - 1.86 * x

Substituting x = [(5/
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Direction: Read the passage given below and answer the following questions: Boiling point or freezing point of liquid solution would be affected by the dissolved solids in the liquid phase. A soluble solid in solution has the effect of raising its boiling point and depressing its freezing point. The addition of non-volatile substances to a solvent decreases the vapour pressure and the added solute particles affect the formation of pure solvent crystals. According to many researches the decrease in freezing point directly correlated to the concentration of solutes dissolved in the solvent. This phenomenon is expressed as freezing point depression and it is useful for several applications such as freeze concentration of liquid food and to find the molar mass of an unknown solute in the solution. Freeze concentration is a high quality liquid food concentration method where water is removed by forming ice crystals. This is done by cooling the liquid food below the freezing point of the solution. The freezing point depression is referred as a colligative property and it is proportional to the molar concentration of the solution (m), along with vapour pressure lowering, boiling point elevation, and osmotic pressure. These are physical characteristics of solutions that depend only on the identity of the solvent and the concentration of the solute. The characters are not depending on the solute’s identity. Q.When a non volatile solid is added to pure water it will

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A 5% solution (w/W) of cane sugar (molar mass = 342 g mol-1)has freezing point of 271 K. What will bethe freezing point of a 5% glucose (molar mass = 18 g mol-1) in water if freezing point of pure water is 273.15 K?a)273.07 Kb)269.07 Kc)273.15 Kd)260.09 KCorrect answer is option 'B'. Can you explain this answer?
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