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MHT CET · Chemistry · Some Basic Concepts of Chemistry

What is the volume occupied by 1 molecule of water, if its density is \(1 \mathrm{~g} \mathrm{~cm}^{-3}\) ?

  1. A \(9.0 \times 10^{-23} \mathrm{~cm}^3\)
  2. B \(2.98 \times 10^{-23} \mathrm{~cm}^3\)
  3. C \(6.023 \times 10^{-23} \mathrm{~cm}^3\)
  4. D \(5.50 \times 10^{-23} \mathrm{~cm}^3\)
Verified Solution

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
\(=\frac{\text { mass }}{\text { density }}=\frac{2.98 \times 10^{-23} \mathrm{~g}}{1 \mathrm{~g} \mathrm{~cm}^{-3}}=2.98 \times 10^{-23} \mathrm{~cm}^3\)