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 0.01 ml is (Density of water is 1 g/ml)

• A. 2.5 X 1016
• B. 2.5 X 1017
• C. 5 X 1016
• D. 7.5 X 1016

Answer 1

0.01 ml 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.
Thus, 0.01g of ordinary water corresponds to $\frac{1}{6000}\times$ 0.01 =

0.16 $\times {10}^{-5}$ g of heavy water.
Molar mass of $D_{2}O$ = 20g
Moles of heavy water = $\frac{\text{givenmass of heavy water}}{\text{molar mass of heavy water}}$ =$\frac{0.16\times {10}^{-5}}{20}=0.008\times {10}^{-5}$
1 mole of heavy water contains = 6.023$\times {10}^{23}$
0.08$\times {10}^{-5}$mole of heavy water contains $=6.023\times {10}^{23}\times 0.008\times {10}^{-5}=4.8\times {10}^{16}=5\times {10}^{16}$
Hence the correct option is C.