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Question

(C) The number of milliequivalents $(Meq.)$ in the concentrated solution is $1600 	imes 0.2050 = 328$.Let the final volume after dilution be $V \, mL

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) The number of milliequivalents $(Meq.)$ in the concentrated solution is $1600 	imes 0.2050 = 328$.Let the final volume after dilution be $V \, mL$.The number of milliequivalents in the diluted solution is $0.20 	imes V$.Since the number of milliequivalents remains constant during dilution,we have $328 = 0.20 	imes V$.Solving for $V$,we get $V = \frac{328}{0.20} = 1640 \, mL$.The volume of water to be added is the difference between the final volume and the initial volume: $1640 \, mL - 1600 \, mL = 40 \, mL$.

What volume of water in $mL$ is required to prepare a $0.20\,N$ solution from $1600\,mL$ of a $0.2050\,N$ solution?

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