The amount of glucose required to prepare 250 mL of 20 M aquepus solution is : (Molar mass of glucose : 180 g mol − 1 )NEET 2024 (Re-NEET)

Molarity, M = M 2 × ( V ) w 2 × 1000 w 2 = Amount of glucose Given molarity = 20 M 20 1 = 180 × 250 w 2 × 1000 w 2 = 20 × 1000 180 × 250 = 2.25 g

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ChemistryLiquid SolutionMiscellaneous/MixedMedium2 minPYQ_2024

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The amount of glucose required to prepare 250 mL of $\frac{M}{20}$ aquepus solution is : (Molar mass of glucose : $180 g mol^{- 1}$ )NEET 2024 (Re-NEET)

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Solution:

Molarity, $M = \frac{w _{2} \times 1000}{M _{2} \times ( V )}$ $w_{2} =$ Amount of glucose Given molarity $= \frac{M}{20}$ $\frac{1}{20} = \frac{w _{2} \times 1000}{180 \times 250}$ $w_{2} = \frac{180 \times 250}{20 \times 1000}$ $= 2.25 g$

Stream:NEETSubject:ChemistryTopic:Liquid SolutionSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2024

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