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KCET · Chemistry · Some Basic Concepts of Chemistry

The number of water molecules present in a drop of water weighing \(0.018 \mathrm{~g}\) is

  1. A \(6.022 \times 10^{26}\)
  2. B \(6.022 \times 10^{23}\)
  3. C \(6.022 \times 10^{19}\)
  4. D \(6.022 \times 10^{20}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(6.022 \times 10^{20}\)

Step-by-step Solution

Detailed explanation

\(18 \mathrm{~g}_{\text {of } \mathrm{H}_{2} \mathrm{O}}=1 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}=6.022 \times 10^{23}\) molecules \(0.018 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}=\frac{0.018}{18} \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}\) \(=\frac{0.018}{18} \times 6.022 \times 10^{23}\) molecules \(=6.022 \times 10^{20}\) molecules