Question

A certain amount of NaOH is dissolved in certain kilograms of solvent and molality of the solution is 0.5 m. When the same amount of NaOH is dissolved in 100 grams of less solvent than initial then molality becomes 0.625 m. Determine the amount of NaOH and the initial mass of solvent.​

Answer 1

$\sf\bold{❍ Given:-}$

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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