Question

Which of the following solution has highest amount of solute?

• A. 1.0 L of 0.25 M \( \text{Na}_2\text{CO}_3 \)
• B. 0.25 L of 0.2 M \( \text{Na}_2\text{SO}_4 \)
• C. 0.5 L of 1.0 M \( \text{KMnO}_4 \)
• D. 0.75 L of 0.5 M \( (\text{NH}_2)_2\text{CO} \)

Hint:

Start by calculating moles (\( M \times V \)). Often the option with the highest moles also has the highest mass unless molar masses are drastically different.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

"Amount of solute" usually refers to the mass or number of moles. Since different compounds are listed, we calculate moles first. If the question implies mass, we'd multiply by molar mass. Usually, in such context, moles are compared, or the product \( M \times V \) is the key.
Step 3: Detailed Explanation:

Calculate moles (\( n = M \times V \)): (A) \( n = 0.25 \times 1.0 = 0.25 \) mol. Mass \( \text{Na}_2\text{CO}_3 \) (106 g/mol) = \( 0.25 \times 106 = 26.5 \) g. (B) \( n = 0.2 \times 0.25 = 0.05 \) mol. Mass \( \text{Na}_2\text{SO}_4 \) (142 g/mol) = \( 0.05 \times 142 = 7.1 \) g. (C) \( n = 1.0 \times 0.5 = 0.5 \) mol. Mass \( \text{KMnO}_4 \) (158 g/mol) = \( 0.5 \times 158 = 79 \) g. (D) \( n = 0.5 \times 0.75 = 0.375 \) mol. Mass \( \text{Urea} \) (60 g/mol) = \( 0.375 \times 60 = 22.5 \) g. Comparison: Moles: (C) 0.5 \textgreater (D) 0.375 \textgreater (A) 0.25 \textgreater (B) 0.05. Mass: (C) 79g \textgreater others. Option (C) has the highest amount (both in moles and mass).
Step 4: Final Answer:

0.5 L of 1.0 M \( \text{KMnO}_4 \) has the highest amount.