Match List I with List II. For preparing 250 mL of aqueous solution w…

Match List I with List II. For preparing \(250 mL\) of aqueous solution with molarity \(0.1\) (List I) the amount of salt required (List II) is:

Choose the correct answer from the options given below:

Answer & explanation

Correct answer: option 2

The correct answer is option 2. A-III, B-IV, C-I, D-II.

A. \(0.1 \text{ M} \text{NaOH}\):

The molar mass of NaOH is approximately 40 g/mol. For 250 mL (0.25 L) of a 0.1 M solution, the required mass is:

\( \text{Mass} = 0.1 \, \text{M} \times 40 \, \text{g/mol} \times 0.25 \, \text{L} = 1 \, \text{g} \)

So, the amount required is 1 g, matching with III.

B. \(0.1 \text{ M} \text{HCl}\):

The molar mass of HCl is approximately 36.46 g/mol. For 250 mL (0.25 L) of a 0.1 M solution, the required mass is:

\(\text{Mass} = 0.1 \, \text{M} \times 36.46 \, \text{g/mol} \times 0.25 \, \text{L} = 0.9115 \, \text{g}\)

So, the amount required is approximately 0.9 g, matching with IV

C. \(0.1 \text{ M} \text{NaCl}\):

The molar mass of NaCl is approximately 58.44 g/mol. For 250 mL (0.25 L) of a 0.1 M solution, the required mass is:

\(\text{Mass} = 0.1 \, \text{M} \times 58.44 \, \text{g/mol} \times 0.25 \, \text{L} = 1.46 \, \text{g}\)

So, the amount required is 1.46 g, matching with I

D. \(0.1 \text{ M} \text{KOH}\):

The molar mass of KOH is approximately 56.11 g/mol. For 250 mL (0.25 L) of a 0.1 M solution, the required mass is:

\(\text{Mass} = 0.1 \, \text{M} \times 56.11 \, \text{g/mol} \times 0.25 \, \text{L} = 1.40 \, \text{g}\)

So, the amount required is 1.4 g, matching with II.

Thus, the correct answer is option 2. A-III, B-IV, C-I, D-II.