The answer for the following problem is mentioned below.
Explanation:
Given:
mass of calcium phosphate ([tex]Ca_{3}(PO_{4} )_{2}[/tex] ) = 125.3 grams
We know;
molar mass of calcium phosphate ([tex]Ca_{3}(PO_{4} )_{2}[/tex] ) = (40×3) + 3 (31 +(4×16))
molar mass of calcium phosphate ([tex]Ca_{3}(PO_{4} )_{2}[/tex] ) = 120 + 3(95)
molar mass of calcium phosphate ([tex]Ca_{3}(PO_{4} )_{2}[/tex] ) = 120 +285 = 405 grams
We also know;
No of molecules at STP conditions([tex]N_{A}[/tex]) = 6.023 × 10^23 molecules
To solve:
no of molecules present in the sample(N)
We know;
[tex]\frac{m}{M} =\frac{N} }{}[/tex]N÷[tex]N_{A}[/tex]
[tex]\frac{405}{125.3}[/tex] =[tex]\frac{N}{6.023*10^23}[/tex]
N =(405×6.023 × 10^23) ÷ 125.3
N = 19.3 × 10^23 molecules
Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules
