QJEEM25Ch9908: JEE Main 2025 Chemistry - Class 11 JEEM25Ch9908 Solution: 20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH…

To determine the final concentration of the NaOH solution, we use the formula for mixing solutions :

$M_F = \frac{M_1 \times V_1 + M_2 \times V_2}{V_1 + V_2}$

Where :

$M_1 = 2 \, \text{M}$ and $V_1 = 20 \, \text{mL}$ : Concentration and volume of the first solution.

$M_2 = 0.5 \, \text{M}$ and $V_2 = 400 \, \text{mL}$ : Concentration and volume of the second solution.

Substitute the values into the equation:

$M_F = \frac{2 \times 20 + 0.5 \times 400}{420}$

Calculating each term:

$2 \times 20 = 40$

$0.5 \times 400 = 200$

Add these results:

$M_F = \frac{40 + 200}{420} = \frac{240}{420} \approx 0.571 \, \text{M}$

Convert this to scientific notation as the problem specifies the answer should be in $\times 10^{-2}$ form:

$M_F = 57.1 \times 10^{-2} \, \text{M}$

Rounded to the nearest integer, the final concentration is:

$57$