Answer
$N = 2.03 \times 10 ^{20}$
Work Step by Step
The mass fraction of $^{226}Ra$ in $RaCl_2$ is calculated using the equation
$m_{Ra} = \frac{M_{Ra}}{M_{Ra} + 2M_{Cl}}(m_{Sample})$
Where
$M_{Cl} = 35.453 g/mol $
$ M_{Ra} = 226 g/mol $
$m_{Sample} = 0.10 g$
So,
$m_{Ra} = \frac{226 g/mol }{226 g/mol + 2(35.453 g/mol )}(0.10 g)$
$m_{Ra} = 0.0761g $
The mass of a single $^{226}Ra$ nucleus is
$(226 u)(1.661 \times 10^{– 24} g/u) = 3.75 \times 10^{– 22}
g$
To know the number of $^{226}Ra$ present, the mass of radium in the sample must be divided by the mass of single $^{226}Ra$ nucleus.
$N = \frac{m_{Ra \space (sample)}}{m_{Ra}}$
$N = 2.03 \times 10 ^{20}$
