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Answer 1

Step-by-step explanation:

$\sf 0.2 M \: 200 \: mL \: NaOH \: and \: 0.5 \:M \: 500 \:mL \: NaOH \:are \:mixed \\ \\ \sf We\:know\:that\: by \: Oswald's \: Dilution \: Law \\ \\ \sf {M}_{1}{V}_{1} + {M}_{2}{V}_{2} = MV \\ \\ \sf Where \: M : Molarity \\ \sf V : Volume \\ \\ \sf Here {M}_{1} = 0.2 \:M \: \: \: \:\: \: {V}_{1} = 200\: mL \\ \sf {M}_{2} = 0.5\: M \:\:\:\:\:\:\: {V}_{2} = 500 \:mL \\ \\ \sf Putting\:the\:values \\ \\ \sf \longrightarrow 0.2 × 200 + 0.5 × 500 = 0.29 × V \\ \\ \sf \longrightarrow 40 + 250 = 0.29V \\ \\ \sf \longrightarrow V = \frac{290}{0.29} \\ \\ \sf \longrightarrow V = 1000 mL \\ \\ \sf So\:total\:volume\:is\:1000 \:mL \: out \:of \: which \: 700 \: mL \: is \: NaOH \\ \\ \sf So \: amount \: of \: water \: added = 1000 - 700 = 300 \: mL$

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Answer 2

Step-by-step explanation:

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