NaOH is dissolved in certain kilograms of solvent and molality of solution 0.5m.
Again , same among of NaOH is dissolved in 100 grams of solvent than initial , then molality becomes 0.625m.
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Now lets find the amount of NaOH and the initial mass of solvent.
Let,
- $\sf\small{Initial\:Mass\:of\:solvent=y}$
- $\sf\small{Number\:of\:moles\:NaOH\:dissolved=x}$
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$\sf\bold{ ❍ We\:know,}$
$\sf{Molality(m)=}$ $\sf\dfrac{No.of\:moles\:of\:solute}{No.of\:solvent\:in\:kg}$
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$\sf\bold{Putting\:the\:formula:-}$
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$\sf\huge\underline\bold{ ❍Case:1}$
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$\longmapsto$ $\sf\small{0.5}$ $\sf\dfrac{x}{y}$
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$\longmapsto$ $\sf{0.5y = x }$
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$\longmapsto$ $\sf{multiply\:by\:2→ y = 2x}$
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$\sf\huge\underline\bold{ ❍Case:2}$
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$\longmapsto$ $\sf\small{0.625}$ $\sf\small\dfrac{x}{y}$ = $\sf\dfrac{x}{y=100g}$
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$\longmapsto$ $\sf\small{0.625}$ $\sf\dfrac{x}{y-100/1000kg}$ = $\sf\dfrac{x}{y-0.1kg}$
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$\longmapsto$ $\sf{0.625(y-0.1kg)=x}$
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$\longmapsto$ $\sf{0.625y-0.0625=x}$
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$\sf\small\bold{By\:putting\:the\:value\:of \:"x" we\: get:}$
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$\longmapsto$ $\sf{0.625(2x)-0.0625 = x}$
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$\longmapsto$ $\sf{1.25x 0.0625 = x}$
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$\longmapsto$ $\sf{1.25x - x = 0.0625}$
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$\longmapsto$ $\sf{0.25x = = 0.0625}$
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$\longmapsto$ $\sf\small{x=}$ $\sf\dfrac{0.0625}{0.25}$= $\sf\bold{x=0.25}$
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$\sf{So,y=2(x)=2\times0.25=}$ $\sf\bold{y=0.5}$
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$\sf\small{Initial\:mass\:of\:solvent:0.5kg=500g}$
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$\sf{Now,}$
Amount of NaOH=
$\space$ $\space$ $\space$ $\space$ $\space$ $\sf{=x\times molar\:mass}$
$\space$ $\space$ $\space$ $\space$ $\space$ $\sf{=0.25\times 40=10}$
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$\sf\underline{\underline{ ⚘ Hence,amount\:of\:NaOH=10kg}}$
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