Question

Be sure to answer all parts. The natural abundances of the two stable isotopes of hydrogen (hydrogen and deuterium) are 99.99 percent and 0.01 percent. Assume that water exists as either 1 1 H2O or 2 1 D2O. Calculate the number of D2O molecules in 300.0 mL of water (density = 1.00 g/mL). Enter your answer in scientific notation.

Answer 1

Answer:

The number of D₂O molecules in 300.0 mL of water are 1,003x10²¹

Explanation:

First, you need to know in 300 mL of water how many moles of hydrogen and deuterium are. Then, you must use Avogadro's number to obtain the number of molecules.

In 300 mL of water you have:

$300,0 mL * \frac{1,00g}{1mL} * \frac{1mol}{18.01528 g}$ =

16,65 moles of H₂O and D₂O

As the 0,01 percent of these moles are D₂O:

16,65 mol × 0,01% = 1,665x10⁻³ mol of D₂O

That in number of molecules are:

$1,665x10^{-3}mol*\frac{6,022x10^{23}molecules}{1mol}$ =

1,003x10²¹ molecules of D₂O

I hope it helps!