Solutions

The correct option is (B) : M₂=

For dilute solution, pV=nRT or V=nRT or V= RT or M₂ = RT ​where, = osmotic pressure V =volume of solution n = number of moles of solute m₂=mass of solute M₂= molecular mass of solute

The correct option is (C) : 0.73 M

Mole of 2.5 L of 1M NaOH =2.5×1=2.5

Mole of 3.0 L of 0.5 M NaOH =3.0×0.5=1.5

Total moles of NaOH in solution =2.5+1.5=4.0 (Total volume of solution =2.5+3.0=5.5L)

Thus, M₁ ×V₁=M₂ ×V₂ ​ 4.0=M₂ ×5.5 Molarity of resultant solution = M₂ = ​ =0.73M

At 25°C, density of water = 1 g/ml

1 g = 1 ml

So, 1000 g = 1000 ml

Now, number of moles of water

Molarity of water

Hence, the molarity of pure water is 55.6 M. 

Answer: Option 4: ΔG_mix = 0

Explanation:

In an ideal solution:

• ΔH_mix = 0 → No enthalpy change occurs during mixing.
• ΔU_mix = 0 → Internal energy change upon mixing is zero.
• ΔP = P_obs − P_{calculated by Raoult’s law} = 0 → Ideal solutions obey Raoult's Law throughout the composition range.

However,

• ΔG_mix ≠ 0 — This is the incorrect statement.
• Even in an ideal solution, mixing leads to an increase in entropy (ΔS_mix > 0), so the Gibbs free energy of mixing decreases (ΔG_mix < 0).

Therefore, ΔG_mix = 0 is incorrect for an ideal solution.

Boiling Point Elevation Calculation

The following formula is used to calculate the change in boiling point:

Where: - is the vapor pressure of the pure solvent, - is the vapor pressure of the solution, - is the molality of the solution, - is the molar mass of the solute (in this case, 18 g/mol for water).

Step 1: Solve for (molality)

Substitute the given values into the equation:

Calculating the left-hand side:

Now solve for :

Step 2: Calculate the Boiling Point Elevation

The boiling point elevation is calculated using the formula:

Where is the ebullioscopic constant of the solvent. Given for water:

Step 3: Calculate the Final Boiling Point

The final boiling point is calculated by adding the boiling point elevation to the normal boiling point of the solvent (which is 100°C for water):

Conclusion:

The final boiling point of the solution is .

To find the vapor pressure of the solution at 45°C with a molar ratio of benzene to octane of 3:2, we can use Raoult’s Law. Raoult’s Law states that the vapor pressure of a solution is directly related to the vapor pressures of the pure components and their mole fractions in the solution.

Raoult's Law formula is given as:

where:

• is the vapor pressure of the solution.
• and are the mole fractions of benzene and octane respectively.
• and are the vapor pressures of pure benzene and octane, respectively.

Given:

• Vapor pressure of pure benzene, mm Hg.
• Vapor pressure of pure octane, mm Hg.
• Mole ratio of benzene to octane is 3:2.

First, calculate the total number of moles:

Next, calculate the mole fractions:

• (mole fraction of benzene)
• (mole fraction of octane)

Now, apply Raoult's Law to calculate the vapor pressure of the solution:

Calculate each term separately:

• mm Hg
• mm Hg

Add both contributions to find the total vapor pressure of the solution:

mm\)

Therefore, the correct option for the vapor pressure of the solution at 45°C is 336 mm of Hg.

To determine the decreasing order of osmotic pressures of the given solutions, we need to consider the concept of osmotic pressure, which is a colligative property. Colligative properties depend on the number of solute particles present in a solution, not on the identity or nature of the solute.

1. The osmotic pressure ( ) of a solution is given by the formula: , where:
   • = Van't Hoff factor (number of particles the solute splits into),
   • = Concentration of the solution in moles per liter,
   • = Universal gas constant, and
   • = Temperature in Kelvin.
2. Let's calculate the Van't Hoff factor and concentration for each solution:
   • Glucose ( ):
     • Does not ionize in solution (i = 1).
     • Molar mass of glucose = 180 g/mol.
     • Moles = mol.
     • Concentration = M.
   • Urea ( ):
     • Does not ionize in solution (i = 1).
     • Molar mass of urea = 60 g/mol.
     • Moles = mol.
     • Concentration = M.
   • Sucrose ( ):
     • Does not ionize in solution (i = 1).
     • Molar mass of sucrose = 342 g/mol.
     • Moles = mol.
     • Concentration = M.
3. The osmotic pressure is proportional to the concentration and Van't Hoff factor:
   • For glucose (P₁): 0.2224 M × 1 = 0.2224
   • For urea (P₂): 0.6668 M × 1 = 0.6668
   • For sucrose (P₃): 0.1168 M × 1 = 0.1168
4. Order of osmotic pressures (highest to lowest):
   • P₂ > P₁ > P₃

Therefore, the correct answer is P₂ > P₁ > P₃, which indicates that urea has the highest osmotic pressure due to its higher concentration, followed by glucose, and then sucrose.

To determine the correct amount of solvent in a molal solution, we need to understand the concept of molality. Molality (m) is defined as the number of moles of solute per kilogram of solvent. Mathematically, it is expressed as:

For a 1 molal solution, the molality is 1 mol/kg. Given the problem:

• Moles of solute: 0.5 mol
• Molality of solution: 1 mol/kg

Using the definition of molality, we set up the equation:

1 = 0.5 / (kilograms of solvent)

To solve for the kilograms of solvent:

kilograms of solvent = = 0.5 kg

Since 1 kilogram equals 1000 grams, we convert the kilograms of solvent into grams:

0.5 kg = 500 g

Therefore, the solution contains 500 g of solvent, which matches the option: 500 g of solvent.

Pumice stone is an example of a colloidal system known as a solid sol. In chemistry, a solid sol consists of solid particles dispersed in a solid medium. This type of colloid retains the shape of the solid while incorporating dispersed phase particles, often with unique characteristics such as porosity or lightness, as seen in pumice stones. Thus, the correct classification of pumice stone is a solid sol.

Chloroform (CHCl₃) is a polar organic solvent. The general principle "like dissolves like" applies here. Polar substances tend to dissolve in polar solvents, and nonpolar substances tend to dissolve in nonpolar solvents.

NaCl: Sodium chloride is an ionic compound. Ionic compounds are generally not soluble in organic solvents like chloroform.

CH₃OH (Methanol): Methanol is a polar molecule due to the presence of the highly electronegative oxygen atom and the polar O-H bond. It can form hydrogen bonds, but the overall polarity is less than NaCl. Methanol exhibits some solubility in chloroform.

CH₃CN (Acetonitrile): Acetonitrile is a polar molecule due to the CN bond. It is less polar than methanol and has a smaller dipole moment. It will be more soluble in chloroform than NaCl but less soluble than methanol.

Cyclohexane: Cyclohexane is a nonpolar molecule. Chloroform has a dipole moment, so it's considered a polar solvent, although it's not as polar as water or methanol. Cyclohexane will be the most soluble in chloroform among the given options because chloroform's polarity is relatively low within the spectrum of polar solvents.

Therefore, the increasing order of solubility in chloroform is: NaCl CH₃CN CH₃OH Cyclohexane.