JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
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 _______ \(\times 10^{-2} \mathrm{M}\). (Nearest integer)
- A 50
- B 57
- C 43
- D 63
Answer & Solution
Correct Answer
(B) 57
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & M_F=\frac{M_1 V_1+M_2 V_2}{V_1+V_2} \\ & =\frac{2 \times 20+0.5 \times 400}{420}=0.571 \mathrm{M} \\ & =57.1 \times 10^{-2} \mathrm{M} \\ & =57\end{aligned}\)
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