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Question: A sample of ammonium phosphate \({\left( {N{H_4}} \right)_3}P{O_4}\) contains 3.18 moles of H-atom. ...

A sample of ammonium phosphate (NH4)3PO4{\left( {N{H_4}} \right)_3}P{O_4} contains 3.18 moles of H-atom. The number of moles of OatomO - atom in the sample is:
A) 0.2650.265
B) 0.7950.795
C) 1.061.06
D) 3.183.18

Explanation

Solution

We know that mole is a fundamental unit for measuring the amount of substance. According to the mole concept, Number of atoms or molecules is given as,
Number of atoms and molecules =(NumberofMoles)×(6.022×1023) = \left( {{\rm{Number}}\,{\rm{of}}\,{\rm{Moles}}} \right) \times \left( {6.022 \times {{10}^{23}}} \right)
And the relationship between the gram and the atomic mass unit (amu) is given as,

1 amu = 1 gram6.022×1023\dfrac{\text{1 gram}}{6.022 \times 10^{23}} = 1.66 ×\times 1024^{-24} grams

Complete step-by-step solution:
Given, the sample of ammonium phosphate (NH4)3PO4{\left( {N{H_4}} \right)_3}P{O_4} contains 3.18 moles of H-atom.
Now, we can calculate the number of moles of OatomO - atom in the sample as follows,
149  grams{\rm{149}}\;{\rm{grams}} of (NH4)3PO4{\left( {N{H_4}} \right)_3}P{O_4} contains 12g of HH.

Now, 3.18 moles of H-atom is present in 149×3.1812=39.5  grams  of  (NH4)3PO4\dfrac{{{\rm{149 \times 3}}{\rm{.18}}}}{{{\rm{12}}}} = {\rm{39}}{\rm{.5}}\;{\rm{grams}}\;{\rm{of}}\;{\left( {{\rm{N}}{{\rm{H}}_{\rm{4}}}} \right)_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{4}}}

149  gram{\rm{149}}\;{\rm{gram}} of (NH4)3PO4{\left( {N{H_4}} \right)_3}P{O_4} contains 64×39.5149  grams\dfrac{{64 \times 39.5}}{{149}}\;{\rm{grams}} of Oxygen =16.96  grams = {\rm{16}}{\rm{.96}}\;{\rm{grams}} of Oxygen.

By this we now know that 16.96  grams{\rm{16}}{\rm{.96}}\;{\rm{grams}} of oxygen is present in the given sample.

\therefore Number of moles of oxygen =16.9616=1.06  moles = \dfrac{{16.96}}{{16}} = 1.06\;{\rm{moles}}

Hence, out of the given options, option C is correct.

Additional Information:
Some important formulas of mole concept are:

Mass of one mole of atoms =Ar  in  grams = {\rm{Ar}}\;{\rm{in}}\;{\rm{grams}}

Number of moles of atoms =MassoftheelementingramsRelativeatomicmass,Ar{\rm{ = }}\dfrac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{element}}\,{\rm{in}}\,{\rm{grams}}}}{{{\rm{Relative}}\,{\rm{atomic}}\,{\rm{mass,}}\,{\rm{Ar}}}}

Mass of 1 mole of molecules = Mr in grams

Number of moles of molecules=MassofthesubstanceingramsRelativeatomicmass,Mr{\rm{ = }}\dfrac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{substance}}\,{\rm{in}}\,{\rm{grams}}}}{{{\rm{Relative}}\,{\rm{atomic}}\,{\rm{mass,}}\,{\rm{Mr}}}}
Mass of substance containing 1 mole of particles = Molar mass

Note: The number of atoms in 1 mole is equal to the Avogadro’s number denoted as (NA)\left( {{{\rm{N}}_{\rm{A}}}} \right). The value of Avogadro’s number is 6.022×10236.022 \times {10^{23}}. The molar mass is given as mass divided by mole that is,
Molar  mass{Molar\; mass} = MassMole\dfrac{Mass}{Mole} = g/mol