KCET · Chemistry · Some Basic Concepts of Chemistry
The number of water molecules present in a drop of water weighing \(0.018 \mathrm{~g}\) is
- A \(6.022 \times 10^{26}\)
- B \(6.022 \times 10^{23}\)
- C \(6.022 \times 10^{19}\)
- D \(6.022 \times 10^{20}\)
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
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