Answer
$m_P = 1.1 \times 10^{-6} g $
Work Step by Step
To know the mass of Phosphorus needed, first find the number of Silicon atoms in 1.0 g.
Molar mass of Silicon is 28.086 g/mol. So,
Mass of one silicon atom,
$m_{0,Si} = \frac{M_{Si}}{N_A}$
$m_{0,Si} = \frac{(28.086 g/mol)}{(6.022 \times 10^{23}/mol)}$
$m_{0,Si} = 4.66 \times 10^{-23}g$
1.0 g of Silicon contains $N_{Si}$ number of atoms, so
$N_{Si} = \frac{m_{Si}}{m_{0,Si}} = \frac{1.0 g}{4.66 \times 10^{-23}g} $
$N_{Si} = 2.14 \times 10^{22} $
From the question, the number density is increased by $ 10^6 $
This means number of phosphorus atoms in 1.0 g is
$N_P = \frac{2.14 \times 10^{22}}{1 \times 10^6} $
$N_P = 2.14 \times 10^{16} $
Molar mass of Phosphorus 30.9758 g/mol. Mass of a single Phosphorus atom is
$m_{0,P} = \frac{M_{P}}{N_A}$
$m_{0,P} = \frac{30.9758 g/mol}{6.022 \times 10^{23}/mol}$
$m_{0,P} = 5.14 \times 10^{-23}g$
So the mass of Phosphorus needed to add in 1.0 g Silicon is
$m_p = (2.14 \times 10^{16} ) ( 5.14 \times 10^{-23}g) $
$m_P = 1.1 \times 10^{-6} g $
