Question

A sample of calcium phosphate was found to have a mass of 125.3 g. How many molecules were contained in the sample?

Answer 1

The answer for the following problem is mentioned below.

• Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules.

Step-by-step explanation:

Given:

mass of calcium phosphate (
`Ca_(3)(PO_(4) )_(2)` ) = 125.3 grams

We know;

molar mass of calcium phosphate (
`Ca_(3)(PO_(4) )_(2)` ) = (40×3) + 3 (31 +(4×16))

molar mass of calcium phosphate (
`Ca_(3)(PO_(4) )_(2)` ) = 120 + 3(95)

molar mass of calcium phosphate (
`Ca_(3)(PO_(4) )_(2)` ) = 120 +285 = 405 grams

We also know;

No of molecules at STP conditions(
`N_(A)`) = 6.023 × 10^23 molecules

To solve:

no of molecules present in the sample(N)

We know;

`(m)/(M) =\frac{N} }{}`N÷
`N_(A)`

`(405)/(125.3)` =
`(N)/(6.023*10^23)`

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