Question

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

• A. $2\times 0.5\times {10}^{16}$
• B. $2\times 0.5\times {10}^{17}$
• C. $5\times {10}^{16}$
• D. $7.5\times {10}^{16}$

Answer 1

The correct option is C $5\times {10}^{16}$

$0.01$ml is equal to $0.01$g of water.
$6000$g of ordinary water corresponds to $1$g of heavy water.
$1$g of ordinary water corresponds to $\frac{1}{6000}$g of heavy water.
Thus, $0.01$g of ordinary water corresponds to $\frac{1\times {10}^{-5}}{6}$ g of heavy water.

Moles of heavy water $=\frac{Weightofheavywater}{Molarmassofheavywater}=\frac{{10}^{-5}}{6\times 20}=\frac{{10}^{-6}}{12}$

No. of molecules $=\frac{6.023\times {10}^{23}\times {10}^{-6}}{12}\approx 5\times {10}^{16}$