The amount of glucose required to prepare 250 mL of $\frac{\mathrm{M}}{20}$ aquepus solution is : (Molar…

The amount of glucose required to prepare 250 mL of $\frac{\mathrm{M}}{20}$ aquepus solution is : (Molar mass of glucose : $180 \mathrm{~g} \mathrm{~mol}^{-1}$ )
  1. 2.25 g
  2. 4.5 g
  3. 0.44 g
  4. 1.125 g

Solution

Molarity, $\mathrm{M}=\frac{\mathrm{w}_2 \times 1000}{\mathrm{M}_2 \times(\mathrm{V})}$ $\mathrm{w}_2=$ Amount of glucose Given molarity $=\frac{\mathrm{M}}{20}$ $\frac{1}{20}=\frac{w_2 \times 1000}{180 \times 250}$ $w_2=\frac{180 \times 250}{20 \times 1000}$ $=2.25 \mathrm{~g}$

Asked in: NEET 2024 (Re-NEET)

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