Properties such as boiling point, freezing point, and vapour pressure of a pure solvent change when solute molecules are added to get homogenous solution. These are called colligative properties.
Anwer the following questions:
i.0.1 M glucose
ii.1% urea solution
iii.0.1 M common salt
Increasing order of osmotic pressure
Properties such as boiling point, freezing point, and vapour pressure of a pure solvent change when solute molecules are added to get homogenous solution. These are called colligative properties.
Anwer the following questions:
i.0.1 M glucose
ii.1% urea solution
iii.0.1 M common salt
Increasing order of osmotic pressure
Anwer the following questions:
i.0.1 M glucose
ii.1% urea solution
iii.0.1 M common salt
Increasing order of osmotic pressure
A
`(i) lt (iii) lt (ii)`
B
`(i) lt (ii) lt (iii)`
C
`(ii) lt (iii) lt (i)`
D
`(iii) lt (i) lt (ii)`
Text Solution
AI Generated Solution
The correct Answer is:
To determine the increasing order of osmotic pressure for the given solutions, we will use the formula for osmotic pressure (π):
\[
\pi = iCRT
\]
Where:
- \( \pi \) = osmotic pressure
- \( i \) = van 't Hoff factor (number of particles the solute dissociates into)
- \( C \) = molarity of the solution
- \( R \) = universal gas constant (which remains constant for our comparison)
Now, let's analyze each solution step by step.
### Step 1: Calculate the osmotic pressure for 0.1 M glucose (i = 1)
- **Concentration (C)** = 0.1 M
- **van 't Hoff factor (i)** = 1 (since glucose does not dissociate)
Using the formula:
\[
\pi_{glucose} = i \cdot C \cdot R = 1 \cdot 0.1 \cdot R = 0.1R
\]
### Step 2: Calculate the osmotic pressure for 1% urea solution
1. **Convert 1% urea solution to molarity**:
- 1% solution means 1 g of urea in 100 mL of solution.
- Molecular weight of urea (NH₂CONH₂) = 60 g/mol.
- Moles of urea = \( \frac{1 \text{ g}}{60 \text{ g/mol}} = \frac{1}{60} \text{ mol} \).
- Molarity (C) = \( \frac{1/60 \text{ mol}}{0.1 \text{ L}} = \frac{1}{6} \text{ M} \).
2. **van 't Hoff factor (i)** = 1 (since urea does not dissociate).
Using the formula:
\[
\pi_{urea} = i \cdot C \cdot R = 1 \cdot \frac{1}{6} \cdot R = \frac{1}{6}R
\]
### Step 3: Calculate the osmotic pressure for 0.1 M common salt (NaCl)
- **Concentration (C)** = 0.1 M
- **van 't Hoff factor (i)** = 2 (since NaCl dissociates into Na⁺ and Cl⁻).
Using the formula:
\[
\pi_{NaCl} = i \cdot C \cdot R = 2 \cdot 0.1 \cdot R = 0.2R
\]
### Step 4: Compare the osmotic pressures
Now we have:
1. \( \pi_{glucose} = 0.1R \)
2. \( \pi_{urea} = \frac{1}{6}R \approx 0.1667R \)
3. \( \pi_{NaCl} = 0.2R \)
### Step 5: Arrange in increasing order
- \( \pi_{glucose} = 0.1R \)
- \( \pi_{urea} \approx 0.1667R \)
- \( \pi_{NaCl} = 0.2R \)
Thus, the increasing order of osmotic pressure is:
\[
\text{0.1 M glucose} < \text{1% urea solution} < \text{0.1 M common salt}
\]
### Final Answer:
**Increasing order of osmotic pressure:**
1. 0.1 M glucose
2. 1% urea solution
3. 0.1 M common salt
To determine the increasing order of osmotic pressure for the given solutions, we will use the formula for osmotic pressure (π):
\[
\pi = iCRT
\]
Where:
- \( \pi \) = osmotic pressure
...
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Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing ethanol and water. Thus moel fraction of ethanol in the mixture is 0.9 . Given: Freezing point depression constant of water (K_(f)^("water")) = 1.86 K kg mol^(-1) Freezing point depression constant of ethanol (K_(f)^("ethanol")) = 2.0 K kg mol^(-1) Boiling point elevation constant of water (K_(b)^("water")) = 0.52 K kg mol^(-1) Boiling point elevation constant of ethanol (K_(b)^("ethanol")) = 1.2 K kg mol^(-1) Standard freezing point of water = 273 K Standard freezing point of ethanol = 155.7K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecualr weight of water = 18 g mol^(-1) Molecular weight of ethanol = 46 g mol^(-1) In asweering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. The vapour pressure of the solution M is:
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. one of its examples is the use of ethylene glycol adn water mixtures as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9 . Given: Freezing point depression constant of water (K_(f)^(water)) = 1.86 K kg mol^(-1) Freezing point depression constant of ethanol (K_(f)^("ethanol")) = 2.0 K kg mol^(-1) Boiling point elevation constant of water (K_(b)^(water)) = 0.52 K kg mol^(-1) Boiling point elevation constant of ethanol (K_(b)^("ethanol")) = 1.2 K kg mol^(-1) Standard freezing point of water = 273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecular weight of water =18 g mol^(-1) Molecular weight of ethanol = 46 g mol^(-1) In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. The vapour pressure of the solution M is
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