Question

15. A 22.4-L sample of which of the following substances, at STP, would contain 6.02 x 10^23 representative particles? *

Answer 1

The answer for 15 is oxygen or carbon dioxide they are the same thing!

The answer for 16 is the formula of the compound and the atomic mas of its elements!

This is for if you take connections!

Answer 2

Answer :

Part 15 : Only gold contains $6.022\times 10^{23}$ number of particles.

Part 16 : The correct option is, (a) the formula of the compound and the atomic mass of its elements.

Explanation :

Part 15 :

As we know that, 1 mole of substance occupies 22.4 L volume of gas and 1 mole of substance contains $6.022\times 10^{23}$ number of particles.

So, 22.4 L volume of gas contains $6.022\times 10^{23}$ number of particles.

(a) Gold :

Gold contains $6.022\times 10^{23}$ number of particles.

(b) Sodium chloride (NaCl) :

In sodium chloride, there are 2 atoms. So, it contains $2\times 6.022\times 10^{23}$ number of particles.

(c) Sulfur (S₈) :

In sulfur, there are 8 atoms. So, it contains $8\times 6.022\times 10^{23}$ number of particles.

(d) Carbon dioxide (CO₂) :

In carbon dioxide, there are 3 atoms. So, it contains $3\times 6.022\times 10^{23}$ number of particles.

Hence, from this we conclude that only gold contains $6.022\times 10^{23}$ number of particles.

Part 16 :

Percent composition : It is calculated by dividing the mass of each element in one mole of the compound that means dividing the mass of each element by the total molar mass of the compound.

Formula used :

$\%\text{ Composition}=\frac{\text{Mass of each element}}{\text{Molar mass of compound}}\times 100$

Hence, the information needed to calculate the percent composition of a compound is, the formula of the compound and the atomic mass of its elements.