Sure, let's go through the problem step by step.
### 6.1.1 Define the term concentration of a solution.
The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent or solution. It is commonly expressed in moles per liter (mol/L), which can also be referred to as molarity (M).
### 6.1.2 Calculate the number of moles of [tex]$HCl( aq )$[/tex] added to sodium thiosulphate.
We are given:
- The concentration of the hydrochloric acid solution, [tex]\( HCl \)[/tex], is [tex]\( 0.2 \, \text{mol/dm}^3 \)[/tex].
- The volume of the hydrochloric acid solution is [tex]\( 200 \, \text{cm}^3 \)[/tex].
First, we need to convert the volume from [tex]\(\text{cm}^3\)[/tex] to [tex]\(\text{dm}^3\)[/tex]:
[tex]\[
1 \, \text{dm}^3 = 1000 \, \text{cm}^3
\][/tex]
[tex]\[
\text{Volume of } HCl = 200 \, \text{cm}^3 \times \frac{1 \, \text{dm}^3}{1000 \, \text{cm}^3} = 0.2 \, \text{dm}^3
\][/tex]
We can now calculate the number of moles of [tex]\( HCl \)[/tex]:
[tex]\[
\text{Number of moles of } HCl = \text{Concentration of } HCl \times \text{Volume of } HCl
\][/tex]
[tex]\[
\text{Number of moles of } HCl = 0.2 \, \text{mol/dm}^3 \times 0.2 \, \text{dm}^3 = 0.04 \, \text{mol}
\][/tex]
### Result:
The number of moles of [tex]\( HCl \)[/tex] added to the sodium thiosulphate is [tex]\( 0.04 \)[/tex] mol.
### 6.1.3 Calculate the volume of [tex]\(SO _2( g )\)[/tex] that will be formed if the reaction takes place at STP.
From the balanced equation, we know that [tex]\( 2 \)[/tex] moles of [tex]\( HCl \)[/tex] produce [tex]\( 1 \)[/tex] mole of [tex]\( SO_2 \)[/tex]:
[tex]\[
2 \, HCl(aq) \rightarrow 1 \, SO_2(g)
\][/tex]
So, the moles of [tex]\( SO_2 \)[/tex] produced can be calculated by:
[tex]\[
\text{Moles of } SO_2 = \frac{\text{Moles of } HCl}{2}
\][/tex]
[tex]\[
\text{Moles of } SO_2 = \frac{0.04 \, \text{mol}}{2} = 0.02 \, \text{mol}
\][/tex]
At Standard Temperature and Pressure (STP), 1 mole of a gas occupies [tex]\( 22.4 \, \text{dm}^3 \)[/tex]. Therefore, the volume of [tex]\( SO_2 \)[/tex] produced can be calculated by:
[tex]\[
\text{Volume of } SO_2 = \text{Moles of } SO_2 \times 22.4 \, \text{dm}^3/\text{mol}
\][/tex]
[tex]\[
\text{Volume of } SO_2 = 0.02 \, \text{mol} \times 22.4 \, \text{dm}^3/\text{mol} = 0.448 \, \text{dm}^3
\][/tex]
To convert [tex]\(\text{dm}^3\)[/tex] to [tex]\(\text{cm}^3\)[/tex]:
[tex]\[
\text{Volume of } SO_2 = 0.448 \, \text{dm}^3 \times 1000 \, \text{cm}^3/\text{dm}^3 = 448 \, \text{cm}^3
\][/tex]
### Result:
The volume of [tex]\( SO_2 \)[/tex] gas formed at STP is [tex]\( 448 \, \text{cm}^3 \)[/tex].
