Answer : Mass of diphosphorous pentoxide will be 1.55 X [tex]10^{-13}[/tex] g.
Explanation : We know that the number of oxygen atoms are, N(O) = 1 X [tex]10^{9}[/tex].
So we have to find the mass of ([tex] P_{2} O_{5} [/tex]) = ?
Now, the number of oxygen atoms is given in atoms so we have to divide it by avogadro's number(Na);
n(O) = N(O) / Na.
On substituting we get, n(O) = 1 [tex]10^{9}[/tex] / 6.022 X [tex]10^{23}[/tex] /mol.
We get, n(O) = 1.66 X [tex]10^{-15}[/tex] mol.
We can take from molecular formula of diphosphorus pentoxide:
number of oxygen and phosphorus ratio;
n(O) : n(P) = 2 : 5.
n(P) = 1.66 X [tex]10^{-15}[/tex] mol X 5/ 2.
So the number of phosphorrus will be n(P) = 4.15 X [tex]10^{-15}[/tex] mol.
Now,m(O) =1.66 X [tex]10^{-15}[/tex] mol X 16 g/mol = 2.65 X [tex]10^{-14}[/tex] g.
and,m(P) = 4.15 X [tex]10^{-15}[/tex] mol X 30.97 g/mol = 1.285 X [tex]10^{-13}[/tex] g.
We can conclude that, m(P₂O₅) = m(O) + m(P).
So, m(P₂O₅) = 2.65 X [tex]10^{-14}[/tex] g + 1.285 X [tex]10^{-13}[/tex] g = 1.55 X [tex]10^{-13}[/tex] g.
Answer:
4.71 × 10⁻¹⁴ g
Solution:
As we know,
141.93 g (1 mole) P₂O₅ contains = 6.022 × 10²³ (5) O atoms
Or,
141.93 g (1 mole) P₂O₅ contains = 3.011 × 10²⁴ O atoms
Therefore, we can say,
3.011 × 10²⁴ O atoms are contained by = 141.93 g of P₂O₅
So,
1.0 × 10⁹ (1 billion) O atoms will be contained by = X g of P₂O₅
Solving for X,
X = (1.0 × 10⁹ × 141.93 g) / 3.011 × 10²⁴
X = 4.71 × 10⁻¹⁴ g
