JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Some Basic Concepts of Chemistry

20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH solution. The final concentration of the solution is ____

\times 1 0^{- 2}

M. (Nearest integer).

Answer: 57

Solution

When two solutions of the same solute are mixed, the moles simply add and so do the volumes, so the final concentration is a volume-weighted average:

M_{F} = \frac{M _{1} V _{1} + M _{2} V _{2}}{V _{1} + V _{2}}

With

M_{1} = 2 M

V_{1} = 20 mL

M_{2} = 0.5 M

V_{2} = 400 mL

M_{F} = \frac{( 2 ) ( 20 ) + ( 0.5 ) ( 400 )}{20 + 400} = \frac{40 + 200}{420} = \frac{240}{420} = 0.5714 M

Expressed as required,

0.5714 M = 57.14 \times 1 0^{- 2} M

, which rounds to 57.

The final concentration of the solution is

57 \times 1 0^{- 2} M

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Integer answer