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

Question

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 $\_\_\_\_$ $	imes 10^{-2} \mathrm{M}$. (Nearest integer)

Professor PC Thomas Classes · Answer

(57) : By molarity equation :\ $M_{1} V_{1}+M_{2} V_{2}=M_{3} V_{3}$\ or, $2 	imes 20+0.5 	imes 400=M_{3} 	imes 420$\ or $40+200=M_{3} 	imes 420$\ or $M_{3}=\frac{240}{420}=57 	imes 10^{-2} \mathrm{M}$