MHT CET · Chemistry · Some Basic Concepts of Chemistry
What is the volume occupied by 1 molecule of water, if its density is \(1 \mathrm{~g} \mathrm{~cm}^{-3}\) ?
- A \(9.0 \times 10^{-23} \mathrm{~cm}^3\)
- B \(2.98 \times 10^{-23} \mathrm{~cm}^3\)
- C \(6.023 \times 10^{-23} \mathrm{~cm}^3\)
- D \(5.50 \times 10^{-23} \mathrm{~cm}^3\)
Answer & Solution
Correct Answer
(B) \(2.98 \times 10^{-23} \mathrm{~cm}^3\)
Step-by-step Solution
Detailed explanation
Mass of \(6.022 \times 10^{23}\) molecules of water \(=1.8 \mathrm{~g}\)
\(\therefore \quad\) Mass of 1 molecule of water
\(\begin{aligned}
& =\frac{18 \mathrm{~g} \times 1}{6.022 \times 10^{23}}=2.98 \times 10^{-23} \mathrm{~g} \\
& \text { Density }=\frac{\text { mass }}{\text { volume }}
\end{aligned}\)
\(\therefore \quad\) Volume occupied by 1 molecule of water
\(=\frac{\text { mass }}{\text { density }}=\frac{2.98 \times 10^{-23} \mathrm{~g}}{1 \mathrm{~g} \mathrm{~cm}^{-3}}=2.98 \times 10^{-23} \mathrm{~cm}^3\)
\(\therefore \quad\) Mass of 1 molecule of water
\(\begin{aligned}
& =\frac{18 \mathrm{~g} \times 1}{6.022 \times 10^{23}}=2.98 \times 10^{-23} \mathrm{~g} \\
& \text { Density }=\frac{\text { mass }}{\text { volume }}
\end{aligned}\)
\(\therefore \quad\) Volume occupied by 1 molecule of water
\(=\frac{\text { mass }}{\text { density }}=\frac{2.98 \times 10^{-23} \mathrm{~g}}{1 \mathrm{~g} \mathrm{~cm}^{-3}}=2.98 \times 10^{-23} \mathrm{~cm}^3\)
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