Chemistry - Concentration Terms Question with Solution | TestHub

ChemistryConcentration TermsConcentration termMedium2 minQB

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Passage / Comprehension

25 molal aqueous solution of a nonvolatile solute (molar mass = $40/3$ g/mol) is partially dried, and it is found that 50% by mass of water has been removed.

% by mass of $H_{2} O$ present in partially dried solution is :

Options:

Answer:

Solution:

Initial Mass: 25 molal means 25 moles of solute ( $25 \times \frac{40}{3} g = \frac{1000}{3} g$ ) in $1000 g$ of water.

After Drying: 50% water is removed, leaving $500 g$ of water while the solute mass remains $\frac{1000}{3} g$ .

Total New Mass: The final solution mass is $500 + \frac{1000}{3} = \frac{2500}{3} g$ .

Percentage of H₂O: $(\frac{500}{2500/3}) \times 100 = \frac{1500}{2500} \times 100 = 60%$

Stream:JEESubject:ChemistryTopic:Concentration TermsSubtopic:Concentration term

⏱ 2mℹ️ Source: QB

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