200 , mL of 0.6 , N H_2SO_4 and 100 , mL of 0 | vedclass

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Question

(A) The formula for mixing two solutions of the same nature is $N_1V_1 + N_2V_2 = N_3V_3$.Here,$N_1 = 0.6 \, N$,$V_1 = 200 \, mL$,$N_2 = 0.3 \, N$,and

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(A) The formula for mixing two solutions of the same nature is $N_1V_1 + N_2V_2 = N_3V_3$.Here,$N_1 = 0.6 \, N$,$V_1 = 200 \, mL$,$N_2 = 0.3 \, N$,and $V_2 = 100 \, mL$.The total volume of the resultant solution is $V_3 = V_1 + V_2 = 200 \, mL + 100 \, mL = 300 \, mL$.Substituting the values into the formula: $(0.6 	imes 200) + (0.3 	imes 100) = N_3 	imes 300$.$120 + 30 = N_3 	imes 300$.$150 = N_3 	imes 300$.$N_3 = \frac{150}{300} = 0.5 \, N$.

$200 \, mL$ of $0.6 \, N$ $H_2SO_4$ and $100 \, mL$ of $0.3 \, N$ $HCl$ are mixed together. The acidic normality of the resultant solution is $........ N$.

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