Solutions

To solve this problem, we need to determine the strength (percentage concentration by weight/volume) of the solution when it is given as an volume solution.

The term "volume strength" refers to the volume of oxygen gas (at standard temperature and pressure) that can be liberated from one volume of hydrogen peroxide solution. In this problem, an volume solution of means that of this solution releases of oxygen gas on decomposition.

The balanced equation for the decomposition of hydrogen peroxide is:

From this equation, we can infer that of produces of at standard temperature and pressure because , and the molar volume of a gas is at STP.

Since of is produced by of the solution, we can write:

Step 1: Calculate the weight of required to produce this volume:

This implies that there are of in of the solution.

Step 2: Calculate the percentage strength of the solution:

The solution has of in , thus the percentage weight/volume ( ) is:

Therefore, the strength of the volume solution of is .

Conclusion: Option is the correct answer.

To determine the weight of benzoic acid required, we start by analyzing the freezing point depression data provided. Given that the molecules dimerize, we need to account for this in our calculations:

1. Understanding Dimerization: When benzoic acid dimerizes, two molecules join to form one dimeric molecule. This affects the colligative properties, such as the depression in freezing point.
2. Relating Freezing Point Depression to Molality:
   • The formula for depression in freezing point is given by: where:
     • (given)
     • (given)
     • is the molality of the solution.
     • is the van't Hoff factor which accounts for the extent of association.
3. Calculating the Van't Hoff Factor:
   • For dimerization, the van't Hoff factor is given by: where is the degree of association.
   • Given (80%), we have: .
4. Substitute values into the freezing point depression formula:
   • Solve for molality : .
5. Calculate the weight of benzoic acid:
   • Molality ( ) is defined as moles of solute per kg of solvent. So, (since 30 g of benzene is used, equivalent to 0.03 kg) where is the molar mass of benzoic acid (122 g/mol).
   • Rearranging gives:
   • Solve for : (rounded to 2.4 g for convenience).

Thus, the correct answer for the weight of benzoic acid that results in the observed depression in freezing point is 2.4 g.

To determine the amount of water separated as ice when the solution is cooled to , we use the concept of freezing point depression. The depression in freezing point is given by the formula:

where:

• is the depression in freezing point.
• is the van 't Hoff factor, which is for non-electrolyte solutions.
• is the cryoscopic constant, for water it is .
• is the molality of the solution.

Step 1: Calculate the molality (m) of ethylene glycol

The molar mass of ethylene glycol ( ) is . Therefore, the molality is calculated as follows:

Step 2: Calculate the freezing point depression

The freezing point depression is calculated using:

The normal freezing point of water is . So, the depressed freezing point is:

Step 3: Determine how much water needs to freeze to achieve a temperature of

Since the given temperature is , which is lower than , some water must separate as ice to achieve this temperature.

The new molality when water separates as ice can be re-calculated:

Let be the mass of water (in kg) that separates as ice:

The new mass of water remaining = kg.

New molality =

At equilibrium (to maintain ):

Solving for :

Thus, the amount of water separated as ice is 64 g.

The correct answer is (C) : (A) – (II), (B) – (III), (C) – (I), (D) – (IV)
(A) Negatively charged sol - CdS sol
(B) Macromolecular colloid - Starch
(C) Positively charged sol - Fe₂O₃.xH₂O
(D) Cheese - A gel

The correct answer is 415 pa
π = i CRT (i = 1)

π = 0.00415 atm
π = 415 Pa

To find the molality of the solution, we need to understand the details provided and make some calculations:

1. First, determine the weight of carbon in the solution. Since the solution has 10.8% carbon by weight in a 250 g solution, the amount of carbon is calculated as: .
2. Next, we need to find the molar mass of -glucose, which is . Calculate its molar mass using atomic weights provided:
   • Carbon ( ):
   • Hydrogen ( ):
   • Oxygen ( ):
3. Now, calculate the moles of carbon in the solution. Since each mole of glucose corresponds to 6 moles of carbon: .
4. For the molality ( ), which is moles of solute per kg of solvent, determine the moles of -glucose based on carbon content:
5. Finally, since glucose is the solute and water is the solvent, the weight of the solvent (water) is calculated as: or .
6. The molality of the solution is given by:

Thus, the molality of the solution is nearest to 2.06.

To find the rise in temperature, we first calculate the moles of HNO₃ and NaOH. Using the formula Moles = Molarity × Volume in L:

Moles of HNO₃ = 0.2 M × 0.6 L = 0.12 mol

Moles of NaOH = 0.1 M × 0.4 L = 0.04 mol

Since NaOH is the limiting reactant, only 0.04 mol of HNO₃ reacts. The heat change (q) from this neutralization is q = −ΔH × moles reacted = −57 kJ/mol × 0.04 mol = −2.28 kJ

This heat is absorbed by the water produced. Mass of water = Total volume = 600 mL + 400 mL = 1000 g (assuming the density of water = 1 g/mL).

Using q = mcΔT, where m = 1000 g, c = 4.2 Jg^–1K^–1, and ΔT is the temperature change:

2.28 kJ = 2280 J = 1000 g × 4.2 Jg^–1K^–1 × ΔT

Solve for ΔT: ΔT = 2280 J / 4200 J/K = 0.5429 K

Expressing the temperature change in the desired form: 0.5429 K = 54.29 × 10^–2°C

Thus, the rise in temperature is 54 × 10^–2°C, fitting the expected range (54,54).

To find the number of millimoles of O₂ that dissolve in 1 litre of water, we employ Henry's Law. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Mathematically, it is expressed as:
P = kH × x
where:

• P is the partial pressure of the gas (0.920 bar).
• kH is the Henry's Law constant (46.82 k bar = 46820 bar, since 1 k bar = 1000 bar).
• x is the mole fraction of the gas in solution.

We rearrange the formula to solve for x:
x = P / kH = 0.920 / 46820

Calculating x gives us:
x ≈ 1.965 × 10^-5

For ideal dilute solutions, the molality is approximately equal to the mole fraction when the solubility is negligible. Thus, we can find the number of moles of O₂ by multiplying x by the number of moles of water in 1 litre. The molar mass of water is approximately 18 g/mol, and 1 litre of water is about 1000 g, equivalent to (1000 g) / (18 g/mol) ≈ 55.56 moles.
Therefore, the number of moles of O₂ is approximately:
moles of O₂ = x × 55.56 ≈ 1.092 × 10^-3

To convert to millimoles, multiply by 1000:
millimoles = 1.092 × 10^-3 × 1000 = 1.092

Rounding to the nearest integer, we find that the number of millimoles of O₂ is:
1

The solubility product of PbS is given as . For the compound PbS, the dissociation in water can be represented as:

Let the solubility of PbS be mol L^–1. Then, the concentration of Pb²⁺ and S^2– ions will both be mol L^–1.

Thus, the expression is given by:

Substituting the value we have:

Taking the square root of both sides gives:

We know:

Using , we get:

Thus, the solubility is:

Therefore, the solubility in the form mol L^–1 is:

Hence, the value of is 282.

The calculated value of falls within the provided range of 282,282, confirming its correctness.

To solve this problem, we employ the ideal gas law and Dalton's Law of Partial Pressures. Given that the total pressure and partial pressure of neon are provided, we can find the partial pressure of molecular oxygen in the mixture.

1. First, use Dalton's Law to find the partial pressure of O₂. Dalton's Law states that the total pressure (P_total) is the sum of the partial pressures of the gases: P_total=P(O₂)+P(Ne).
   • We have P_total=25 bar and P(Ne)=20 bar.
   • Thus, P(O₂)=P_total−P(Ne)=25 bar−20 bar=5 bar.
2. Next, we calculate the moles of Ne using the ideal gas law: PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.
   • We don't need exact values for V, R, and T, since they will cancel out, considering both gases are in the same conditions.
   • For Ne: P(Ne)V=n(Ne)RT implies n(Ne)=P(Ne)V/RT.
   • Given the weight of Ne is 200 g and its molar mass is 20 g/mol, n(Ne)=200 g / 20 g/mol=10 mol.
3. Since we have determined the partial pressure of O₂ and know the relationship with moles, we calculate the moles of O₂ similarly: n(O₂)=P(O₂)V/RT.
4. Using the ratio of moles and pressures (because V, R, and T cancel out), n(O₂)=P(O₂)/P(Ne) * n(Ne)=5 bar / 20 bar * 10 mol=2.5 mol.
5. Convert moles to grams for oxygen: mass of O₂=n(O₂) * molar mass of O₂=2.5 mol * 32 g/mol=80 g.

The value of ‘x’, which is the mass of O₂, is 80 g. It falls within the range 80 to 80, confirming correctness.

The correct answer is 3
ΔT_b = i × K_b × m
ΔT_f = i × K_f × m i = 1
4 = 1 × K_b × 1.5
4 = 1 × K_f × 4.5