Chosse the correct statements

Options

A. 1 M `H_(2)SO_(4)` (d=1 gm/ml) is more concentrated than 1 m `H_(2)SO_(4)` (d=1 gm/ml)
B. Molality of solution is 1.136 if 2 gram-equivalents of `H_(2)SO_(4)`is dissolved into 90.2 gm water
C. Vapour pressure of solution becomes higher than ideal solution if there is a positive deviation (according to Roult's low)
D. When 0.1 M `K_(4)[Fe(CN)_(6)]` solution and 0.1 M `FeCl_(3)` solution is separated by a semipermeable membrane , water flows from `K_(4)[Fe(CN)_(6)]` solution to `FeCl_(3)` solution

Answer

To solve the question of choosing the correct statements regarding solutions, we will analyze each option step by step.

### Step 1: Analyze Option 1
**Statement:** 1M H2SO4 is more concentrated than 1m H2SO4.

**Solution:**
- **Definitions:**
- Molarity (M) is defined as moles of solute per liter of solution.
- Molality (m) is defined as moles of solute per kilogram of solvent.

- **Comparison:**
- For 1M H2SO4, if we consider 1 liter of solution, it contains 1 mole of H2SO4 and the volume of the solvent (water) is less than 1 liter.
- For 1m H2SO4, if we consider 1 mole of H2SO4, it is dissolved in 1 kilogram (1000 grams) of water.

- **Conclusion:**
- Since the volume of the solution in molarity includes both solute and solvent, while in molality, the mass of the solvent is considered without the solute, the concentration of 1M H2SO4 is indeed higher than that of 1m H2SO4.

**Result:** Option 1 is correct.

### Step 2: Analyze Option 2
**Statement:** Molality of a solution is 1.136 if 2 g equivalent of H2SO4 is dissolved into 90.2 g of water.

**Solution:**
- **Given:**
- 2 g equivalents of H2SO4.
- Mass of water = 90.2 g.

- **Calculating Weight of H2SO4:**
- Equivalent weight of H2SO4 = Molecular weight / n-factor = 98 g/mol / 2 = 49 g/equiv.
- Therefore, weight of H2SO4 = 2 equiv × 49 g/equiv = 98 g.

- **Calculating Moles of H2SO4:**
- Moles of H2SO4 = Weight / Molecular weight = 98 g / 98 g/mol = 1 mole.

- **Calculating Molality:**
- Molality (m) = Moles of solute / Mass of solvent (in kg) = 1 mole / (90.2 g / 1000) = 1 / 0.0902 = 11.09 m.

- **Conclusion:**
- The calculated molality is approximately 11.09, which is not equal to 1.136.

**Result:** Option 2 is incorrect.

### Step 3: Analyze Option 3
**Statement:** Vapor pressure of solution becomes higher than the ideal solution if there is positive deviation.

**Solution:**
- **Raoult's Law:** States that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent.
- **Positive Deviation:** Occurs when the interactions between solute and solvent are weaker than those in the pure components, leading to a higher vapor pressure than expected.

**Conclusion:**
- This statement is true as it aligns with the principles of Raoult's Law regarding positive deviation.

**Result:** Option 3 is correct.

### Step 4: Analyze Option 4
**Statement:** Water flows from K4Fe(CN)6 solution to FeCl3 solution.

**Solution:**
- **Dissociation:**
- K4Fe(CN)6 dissociates into 5 particles (4K+ + Fe(CN)6^4-).
- FeCl3 dissociates into 4 particles (Fe^3+ + 3Cl^-).

- **Concentration Comparison:**
- Both solutions have the same molarity (0.1 M), but K4Fe(CN)6 produces more particles, resulting in a higher osmotic pressure.

- **Direction of Water Flow:**
- Water will flow from the solution with lower solute potential (K4Fe(CN)6) to the one with higher solute potential (FeCl3).

**Conclusion:**
- The statement is incorrect as water flows from K4Fe(CN)6 to FeCl3, not the other way around.

**Result:** Option 4 is incorrect.

### Final Results:
- **Correct Options:** 1 and 3.
- **Incorrect Options:** 2 and 4.