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

Correct answer: A. 2.25 g Molarity, M= w_2 × 1000 M_2 ×(V) w_2= Amount of glucose Given molarity = M 20 (1)/(20)=(w_2 × 1000)/(180 × 250) w_2=(180 × 250)/(20 × 1000) =2.25 ~g

NEET 2024

Exam: NEET
Year: 2024
Subject: Chemistry
Chapter: Solutions

NEET 2024 · Chemistry · Solutions

The amount of glucose required to prepare 250 mL of M 20 aquepus...

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}

)

Options

A2.25 g
B4.5 g
C0.44 g
D1.125 g

Correct answer

A. 2.25 g

Step-by-step 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}

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