Question

Ordinary water contains one part of heavy water per $6000$ parts by weight. The number of heavy water molecules present in a drop of water of volume $0.01mL$ is (Density of water is $1g/mL$)

Answer 1

$\because$ Density of water $=1g/mL$
$\therefore$$0.01mL$ is equal to $0.01g$ of water.

$6000g$ of ordinary water corresponds to $1g$ of heavy water.

$1g$ of ordinary water corresponds to $\frac{1}{6000}g$ of heavy water.

$0.001g$ of ordinary water corresponds to $\frac{1\times {10}^{-5}}{6}g$ of heavy water.

$\text{Moles of heavy water}=\frac{\text{Weight of heavy water}}{\text{Molar mass of heavy water}}=\frac{{10}^{-5}}{6\times 20}=\frac{{10}^{-6}}{12}$

Number of heavy water molecules $=\frac{6.023\times {10}^{23}\times {10}^{-6}}{12}=5\times {10}^{16}$

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