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 0.001ml is (density of water is 1g/ml

Answer 1

6000g ------1g
0.01g---------?
1.6×10^-6 g
heavy water ---D20

20g'-----6×10^23
1.6×10^-6 ----

1 part of D2o --- 6000 parts of H2O

6000g D2o -------1g of H2O
x----------------0.01
please find x

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Answer 2

Answer: $48.2\times 10^{14}$ molecules

Explanation: Given : 1 gram of heavy water in 6000 g of ordinary water.

Given : Volume = 0.001 ml

Density of wtaer = 1 g/ml

Thus $mass=density\times Volume=1g/ml\times 0.001ml=0.001 g$

If 6000 g of ordinary water contains 1 gram of heavy water.

0.001 gram of ordinary water will contain=$\frac{1g}{6000g}\times 0.001=1.6\times 10^{-7}g$ of heavy water.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{1.6\times 10^{-7}}{20g/mol}$

$\text{Number of moles}=8\times 10^{-9}$

Now 1 mole of heavy water contains contains = $6.023\times 10^{23}$ molecules

$8\times 10^{-9}moles$ of heavy water contains = $\frac{6.023\times 10^{23}}{1}\times 8\times 10^{-9}=48.2\times 10^{14}$ molecules