Question

Calculate the molality of KI if the density of 20% (mass/mass) aqueous solution of KI is (1)202 g mL$^{-1}$.
(Molar mass of KI is 166 g mol$^{-1}$)

• A. (1)5 mol kg$^{-1}$
• B. (1)2 mol kg$^{-1}$
• C. (1)5 mol kg$^{-1}$
• D. 0.12 mol kg$^{-1}$

Hint:

Remember to use the density of the solution to calculate the mass of the solvent when molality is involved.

Answer 1

The Correct Option is A

To calculate the molality of KI in a 20% (mass/mass) aqueous solution of KI, follow the steps below:

Understand the Definitions:

• The density of the solution is given as \(1.202 \, \text{g mL}^{-1}\).
• A 20% (mass/mass) solution means 20 g of KI is present in every 100 g of solution.
• Molality is defined as the number of moles of solute per kilogram of solvent.

Calculate the Mass of Solvent:

• 20% (w/w) means 20 g KI and 80 g of water (since 100 g - 20 g = 80 g of solvent).

Calculate the Number of Moles of KI:

• The molar mass of KI is given as \(166 \, \text{g mol}^{-1}\).
• The number of moles of KI in 20 g of KI is calculated as: 
  \(\text{Moles of KI} = \frac{20}{166} \approx 0.1205 \, \text{mol}\)

Convert Mass of Solvent to Kilograms:

• Convert 80 g to kg: 
  \(80 \, \text{g} = 0.080 \, \text{kg}\)

Calculate the Molality:

• Molality is calculated using the formula: 
  \(\text{Molality} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}\)
• Substitute the calculated values: 
  \(\text{Molality} = \frac{0.1205}{0.080} \approx 1.506 \, \text{mol kg}^{-1}\)

The closest option to our calculated molality is (1) 5 mol kg^-1, implying either an error in calculation or a misinterpretation of density. However, this value doesn't align, meaning there might be an error in the provided options as correctly calculated value using provided data yields approximately 1.506 mol kg^-1.