20 mL of 2 M NaOH solution is added to 400 | vedclass

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Question

(A) The formula for the final concentration $(M_F)$ of a mixture is $M_F = \frac{M_1 V_1 + M_2 V_2}{V_1 + V_2}$.Given: $M_1 = 2 \ M$,$V_1 = 20 \ mL$,$

Vedclass · Accepted Answer

$57$

Vedclass · Answer

(A) The formula for the final concentration $(M_F)$ of a mixture is $M_F = \frac{M_1 V_1 + M_2 V_2}{V_1 + V_2}$.Given: $M_1 = 2 \ M$,$V_1 = 20 \ mL$,$M_2 = 0.5 \ M$,$V_2 = 400 \ mL$.Substituting the values: $M_F = \frac{2 	imes 20 + 0.5 	imes 400}{20 + 400} = \frac{40 + 200}{420} = \frac{240}{420} \approx 0.5714 \ M$.Converting to the required format: $0.5714 \ M = 57.14 	imes 10^{-2} \ M$.Rounding to the nearest integer,we get $57 	imes 10^{-2} \ M$.

$20 \ mL$ of $2 \ M$ $NaOH$ solution is added to $400 \ mL$ of $0.5 \ M$ $NaOH$ solution. The final concentration of the solution is $.......... \times 10^{-2} \ M$. $(Nearest \ integer)$.

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