A sample of pure potassium trioxonitrate(V), \text{KNO}_3, contains 1.806 \times 10^{23} o… — JAMB UTME

A sample of pure potassium trioxonitrate(V), $\text{KNO}_3$ , contains $1.806 \times 10^{23}$ oxygen atoms. What is the mass, in grams, of this sample of $\text{KNO}_3$ ? $[\text{K} = 39, \text{N} = 14, \text{O} = 16, N_A = 6.02 \times 10^{23}\text{ mol}^{-1}]$

Answer: 10.1 g

Answer

The mass of the potassium trioxonitrate(V) sample is 10.1 g.

First, the number of moles of oxygen atoms is found by dividing

1.806 \times 10^{23}

atoms by Avogadro's number (

6.02 \times 10^{23}\text{ mol}^{-1}

), yielding

0.3\text{ mol}

of oxygen. Since one mole of

\text{KNO}_3

contains three moles of oxygen atoms, the moles of

\text{KNO}_3

in the sample is

0.3 / 3 = 0.1\text{ mol}

. Multiplying

0.1\text{ mol}

by the molar mass of

\text{KNO}_3

(

101\text{ g/mol}

) gives the correct mass of

10.1\text{ g}

Step-by-Step Solution

Calculate the moles of oxygen atoms in the sample

n(\text{O}) = 0.3\text{ mol}

Divide the total number of oxygen atoms by Avogadro's constant (

N_A = 6.02 \times 10^{23}\text{ mol}^{-1}

Determine the moles of potassium trioxonitrate(V),

\text{KNO}_3

n(\text{KNO}_3) = 0.1\text{ mol}

Each formula unit of

\text{KNO}_3

contains 3 oxygen atoms, so divide the moles of oxygen atoms by 3.

Calculate the molar mass of

\text{KNO}_3

M(\text{KNO}_3) = 101\text{ g/mol}

Sum the relative atomic masses:

39\text{ (K)} + 14\text{ (N)} + 3 \times 16\text{ (O)} = 101\text{ g/mol}

Multiply moles of

\text{KNO}_3

by its molar mass to get total mass

\text{Mass} = 10.1\text{ g}

\text{Mass} = \text{moles} \times \text{molar mass} = 0.1\text{ mol} \times 101\text{ g/mol} = 10.1\text{ g}

Key Concept

Relationship between particle count, mole quantity of constituent atoms, and molar mass

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