Question

Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and PQ\(_2\). When 1 g of PQ is dissolved in 50 g of solvent 'A', \(\Delta T_b\) was 1.176 K while when 1 g of PQ\(_2\) is dissolved in 50 g of solvent 'A', \(\Delta T_b\) was 0.689 K. (\(K_b\) of 'A' = 5 K kg mol\(^{-1}\)). The molar masses of elements P and Q (in g mol\(^{-1}\)) respectively, are:

• A. 70, 110
• B. 60, 25
• C. 25, 60
• D. 65, 145

Hint:

In problems involving colligative properties, always ensure your units are consistent. The mass of the solvent must be in kilograms when using the standard K_b or K_f values. Setting up and solving systems of linear equations is a common follow-up step in such questions.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This problem involves using the colligative property of elevation in boiling point (Δ$T_b$) to determine the molar masses of two unknown compounds, PQ and PQ₂. From these molar masses, we need to find the atomic masses of the elements P and Q.
Step 2: Key Formula or Approach:
The formula for elevation in boiling point is:
\[ \Delta T_b = K_b \times m \] where \(m\) is the molality of the solution. Molality is given by:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} = \frac{\text{mass of solute / Molar mass of solute}}{\text{mass of solvent (in kg)}} \] We can rearrange the formula to solve for the Molar mass (M) of the solute:
\[ M = \frac{K_b \times \text{mass of solute}}{\Delta T_b \times \text{mass of solvent (in kg)}} \] Step 3: Detailed Explanation:
Given data:
\begin{itemize} \item Mass of solute (for both PQ and PQ₂) = 1 g
\item Mass of solvent 'A' = 50 g = 0.050 kg
\item Ebullioscopic constant of 'A', \(K_b\) = 5 K kg mol⁻¹
\item For compound PQ, \( \Delta T_b \) = 1.176 K
\item For compound PQ₂, \( \Delta T_b \) = 0.689 K
\end{itemize} Calculation of Molar Mass of PQ (M_PQ):
\[ M_{PQ} = \frac{5 \times 1}{1.176 \times 0.050} = \frac{5}{0.0588} \approx 85.03 \text{ g mol}^{-1} \] Calculation of Molar Mass of PQ₂ (M_PQ₂):
\[ M_{PQ_2} = \frac{5 \times 1}{0.689 \times 0.050} = \frac{5}{0.03445} \approx 145.14 \text{ g mol}^{-1} \] Finding Atomic Masses of P and Q:
Let the atomic mass of element P be \(M_P\) and that of element Q be \(M_Q\).
We have a system of two linear equations:
1) \( M_P + M_Q = 85.03 \)
2) \( M_P + 2 M_Q = 145.14 \)
Subtracting equation (1) from equation (2):
\[ (M_P + 2 M_Q) - (M_P + M_Q) = 145.14 - 85.03 \] \[ M_Q = 60.11 \text{ g mol}^{-1} \approx 60 \text{ g mol}^{-1} \] Substituting the value of \(M_Q\) back into equation (1):
\[ M_P + 60.11 = 85.03 \] \[ M_P = 85.03 - 60.11 = 24.92 \text{ g mol}^{-1} \approx 25 \text{ g mol}^{-1} \] Step 4: Final Answer:
The atomic mass of P is approximately 25 g mol⁻¹ and the atomic mass of Q is approximately 60 g mol⁻¹. This corresponds to option (C).