The volume of water added to 500 mL of 0.5 | vedclass

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Question

(D) Initial volume of $NaOH$ solution = $500 \ mL$ and molarity = $0.5 \ M$.Number of moles of $NaOH = 	ext{Molarity} 	imes 	ext{Volume (in L)} = 0

Vedclass · Accepted Answer

$500$

Vedclass · Answer

(D) Initial volume of $NaOH$ solution = $500 \ mL$ and molarity = $0.5 \ M$.Number of moles of $NaOH = 	ext{Molarity} 	imes 	ext{Volume (in L)} = 0.5 	imes 0.5 = 0.25 \ mol$.Mass of $NaOH = 	ext{Moles} 	imes 	ext{Molar mass} = 0.25 	imes 40 \ g = 10 \ g$.Since $1 \ g = 1000 \ mg$,mass of $NaOH = 10000 \ mg$.Let the volume of water added be $V \ mL$. The final volume of the solution = $(500 + V) \ mL$.Final strength = $\frac{	ext{Mass of solute (mg)}}{	ext{Volume of solution (mL)}} = \frac{10000}{500 + V} = 10 \ mg \ mL^{-1}$.$1000 = 500 + V$.$V = 500 \ mL$.

The volume of water added to $500 \ mL$ of $0.5 \ M \ NaOH$ solution so that its strength becomes $10 \ mg \ mL^{-1}$ will be .......... $mL$.

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