Q.Calculate

Concept: Calculation of solution concentrations (molality, molarity, mole fraction) from mass percentage and density.

Reasoning:

1. Assume 100 g of the solution. This means 20 g of KI (solute) and $100 - 20 = 80$ g of water (solvent).
2. Calculate the moles of KI and water using their molar masses ($M_{KI} = 166 \text{ g/mol}$, $M_{H_2O} = 18 \text{ g/mol}$).
3. Determine the volume of the 100 g solution using the given density.
4. Apply the definitions for molality, molarity, and mole fraction.

Calculations:

• Mass of KI = 20 g
• Mass of $H_2O$ = 80 g $= 0.080$ kg
• Moles of KI ($n_{KI}$) = $\frac{20 \text{ g}}{166 \text{ g/mol}} = 0.12048 \text{ mol}$
• Moles of $H_2O$ ($n_{H_2O}$) = $\frac{80 \text{ g}}{18 \text{ g/mol}} = 4.4444 \text{ mol}$
• Volume of solution ($V$) = $\frac{\text{Mass of solution}}{\text{Density}} = \frac{100 \text{ g}}{1.202 \text{ g/mL}} = 83.19 \text{ mL} = 0.08319 \text{ L}$

1. Molality ($m$) = $\frac{n_{KI}}{\text{Mass of } H_2O \text{ (kg)}} = \frac{0.12048 \text{ mol}}{0.080 \text{ kg}} = 1.506 \approx 1.5 \text{ mol/kg}$
2. Molarity ($M$) = $\frac{n_{KI}}{V \text{ (L)}} = \frac{0.12048 \text{ mol}}{0.08319 \text{ L}} = 1.448 \approx 1.45 \text{ mol/L}$ …

When dealing with concentration terms like molality, molarity, and mole fraction, it's crucial to understand what each term represents and how they relate to the amounts of solute and solvent. The problem provides the concentration as a mass percentage and the density of the solution. This combination allows us to determine all other concentration units.

The key idea is to establish a convenient basis for our calculations. Since we are given a mass percentage, assuming a total mass of the solution (like 100 g or 1 kg) simplifies the initial steps. From this assumed mass, we can directly find the mass of the solute and solvent. The density then becomes essential for converting the total mass of the solution into its volume, which is required for molarity.

Let's define the terms we need to calculate:

• Molality ($m$): Moles of solute per kilogram of solvent. It is independent of temperature.

  $m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}$

• Molarity ($M$): Moles of solute per liter of solution. It is temperature-dependent because volume changes with temperature.

  $M = \frac{\text{moles of solute}}{\text{volume of solution (L)}}$

• Mole Fraction ($\chi$): Moles of a component divided by the total moles of all components in the solution. It is a dimensionless quantity.

  $\chi_{\text{solute}} = \frac{\text{moles of solute}}{\text{total moles of all components}}$

We will use the following molar masses:

• Molar mass of KI ($M_{KI}$) = $39.098 \text{ g/mol (K)} + 126.904 \text{ g/mol (I)} = 166.002 \text{ g/mol}$
• Molar mass of H$_2$O ($M_{H_2O}$) = $2 \times 1.008 \text{ g/mol (H)} + 15.999 \text{ g/mol (O)} = 18.015 \text{ g/mol}$

Now, let's proceed with the calculations step-by-step.

1. Establish a Basis for Calculation

   The solution is 20% (mass/mass) aqueous KI. This means that for every 100 g of solution, there are 20 g of KI.

   Let's assume we have exactly $100 \text{ g}$ of the solution.

   • Mass of KI (solute) = $20\%$ of $100 \text{ g} = 20 \text{ g}$
   • Mass of water (solvent) = Total mass of solution - Mass of KI Mass of water = $100 \text{ g} - 20 \text{ g} = 80 \text{ g}$

2. Calculate Moles of Solute (KI) and Solvent (Water)

   To find molality, molarity, and mole fraction, we need the number of moles of each component.

   • Moles of KI ($n_{KI}$): $n_{KI} = \frac{\text{mass of KI}}{M_{KI}} = \frac{20 \text{ g}}{166.002 \text{ g/mol}} = 0.12048 \text{ mol}$
   • Moles of water ($n_{H_2O}$): $n_{H_2O} = \frac{\text{mass of water}}{M_{H_2O}} = \frac{80 \text{ g}}{18.015 \text{ g/mol}} = 4.4407 \text{ mol}$

3. (a) Calculate Molality ($m$)

   Molality is moles of solute per kilogram of solvent.

   • Mass of solvent (water) in kg = $80 \text{ g} \times \frac{1 \text{ kg}}{1000 \text{ g}} = 0.080 \text{ kg}$
   • $m = \frac{n_{KI}}{\text{mass of water (kg)}} = \frac{0.12048 \text{ mol}}{0.080 \text{ kg}} = 1.506 \text{ mol/kg} \approx 1.5 \text{ mol/kg}$

4. Calculate Volume of Solution

   Molarity requires the volume of the solution. We use the given density.

   • Density of solution = $1.202 \text{ g mL}^{-1}$
   • Mass of solution = $100 \text{ g}$ (from our assumed basis) …

Method: Stepwise Density–Mass–Mole Conversion

We'll use the mass-based definition approach — starting from the given mass percentage, then using density to connect mass and volume.

Step 1: Understand the given data

• 20% (mass/mass) means: 20 g KI per 100 g solution
• Density of solution = $1.202 \, \text{g mL}^{-1}$
• Molar mass of KI: $K = 39.1$, $I = 126.9$ $\Rightarrow M_{\text{KI}} = 39.1 + 126.9 = 166.0 \, \text{g mol}^{-1}$

Step 2: Find molality (m)

Molality = moles of solute per kg of solvent.

• Mass of solute (KI) = $20 \, \text{g}$

• Mass of solution = $100 \, \text{g}$

• Mass of solvent (water) = $100 - 20 = 80 \, \text{g} = 0.080 \, \text{kg}$

• Moles of KI:

• Molality:

Answer (a): $\boxed{1.5 \, \text{mol kg}^{-1}}$

Step 3: Find molarity (M)

Molarity = moles of solute per litre of solution.

• Mass of solution = $100 \, \text{g}$
• Density = $1.202 \, \text{g mL}^{-1}$
• Volume of solution:

• Moles of KI = $0.1205 \, \text{mol}$ (from Step 2)

• Molarity:

Answer (b): $\boxed{1.45 \, \text{mol L}^{-1}}$

Step 4: Find mole fraction of KI

Mole fraction = moles of KI divided by total moles (KI + water).

• Moles of KI = $0.1205 \, \text{mol}$
• Moles of water: Mass of water = $80 \, \text{g}$, molar mass $H_2O = 18.0 \, \text{g mol}^{-1}$ …

Common Mistakes in Molality Calculation (and How to Avoid Them)

This problem from the solution chapter (Class 12 Chemistry, CBSE / JEE / NEET) is a classic trap. Here are the most frequent errors students make — and the concept-first fix for each.

✗ Mistake 1: Confusing molality with molarity

• What students do:

  They use the volume of solution (from density) to calculate molality.

• Why it’s wrong:

  Molality ($m$) is moles of solute per kg of solvent, not per kg of solution.

• How to avoid: Always ask: “Is the denominator the solvent mass or the solution mass?” For molality → solvent mass. For molarity → solution volume.

✗ Mistake 2: Using the wrong mass of solvent

• What students do:

  They take 100 g of solution, find 20 g KI, and then use 100 g as the solvent mass.

• Why it’s wrong:

  In a 20% (mass/mass) solution:

  • Mass of KI (solute) = 20 g
  • Mass of water (solvent) = 100 g – 20 g = 80 g Using 100 g as solvent gives a 25% error.

• How to avoid:

  Write clearly:

✗ Mistake 3: Forgetting to convert g to kg for molality

• What students do:

  They plug 80 g directly into the molality formula.

• Why it’s wrong:

  Molality requires kg of solvent.

• How to avoid: Always convert:

✗ Mistake 4: Using density incorrectly for molarity

• What students do:

  They take density = 1.202 g/mL and directly multiply by 1000 to get “mass of 1 L solution” — but forget that density is for solution, not solvent.

• Why it’s wrong:

  Density helps find volume of solution from mass. For molarity:

Volume is found from:

• How to avoid: Step-by-step:
  1. Assume 100 g solution (since 20% is mass/mass).
  2. Volume = $\frac{100}{1.202}$ mL → convert to L.
  3. Moles of KI = $\frac{20}{\text{Molar mass of KI}}$.

✗ Mistake 5: Mole fraction — forgetting both components

• What students do:

  They calculate only moles of KI and divide by moles of water — but mole fraction requires sum of moles of all components.

• Why it’s wrong: …