Answer :
To determine the volume of hydrogen chloride (HCl) gas at standard temperature and pressure (S.T.P) that would yield a [tex]\(1.2 \, \text{dm}^3\)[/tex] solution with a molarity of [tex]\(6.15 \, \text{M}\)[/tex], we can follow these steps:
1. Calculate the number of moles of HCl in the solution:
Molarity ([tex]\(M\)[/tex]) is defined as the number of moles of solute per liter (or cubic decimeter) of solution. Given the volume of the solution and its molarity, we can calculate the number of moles of HCl using the formula:
[tex]\[ \text{Number of moles of HCl} = \text{Molarity} \times \text{Volume of solution} \][/tex]
Given:
- Volume of solution = [tex]\(1.2 \, \text{dm}^3\)[/tex]
- Molarity of the solution = [tex]\(6.15 \, \text{M}\)[/tex]
[tex]\[ \text{Number of moles of HCl} = 6.15 \, \text{M} \times 1.2 \, \text{dm}^3 = 7.38 \, \text{moles} \][/tex]
2. Calculate the volume of HCl gas at S.T.P:
At standard temperature and pressure (S.T.P), 1 mole of any ideal gas occupies [tex]\(22.4 \, \text{dm}^3\)[/tex] of volume.
To find the volume of [tex]\(7.38\)[/tex] moles of HCl gas, we use the relationship:
[tex]\[ \text{Volume of HCl gas at S.T.P} = \text{Number of moles of HCl} \times \text{Volume of one mole of gas at S.T.P} \][/tex]
Given:
- Number of moles of HCl = [tex]\(7.38\)[/tex]
- Volume of one mole of gas at S.T.P = [tex]\(22.4 \, \text{dm}^3\)[/tex]
[tex]\[ \text{Volume of HCl gas at S.T.P} = 7.38 \, \text{moles} \times 22.4 \, \text{dm}^3/\text{mole} = 165.312 \, \text{dm}^3 \][/tex]
Therefore, the volume of hydrogen chloride gas at S.T.P that would yield a [tex]\(1.2 \, \text{dm}^3\)[/tex] solution with a molarity of [tex]\(6.15 \, \text{M}\)[/tex] is [tex]\(165.312 \, \text{dm}^3\)[/tex].
1. Calculate the number of moles of HCl in the solution:
Molarity ([tex]\(M\)[/tex]) is defined as the number of moles of solute per liter (or cubic decimeter) of solution. Given the volume of the solution and its molarity, we can calculate the number of moles of HCl using the formula:
[tex]\[ \text{Number of moles of HCl} = \text{Molarity} \times \text{Volume of solution} \][/tex]
Given:
- Volume of solution = [tex]\(1.2 \, \text{dm}^3\)[/tex]
- Molarity of the solution = [tex]\(6.15 \, \text{M}\)[/tex]
[tex]\[ \text{Number of moles of HCl} = 6.15 \, \text{M} \times 1.2 \, \text{dm}^3 = 7.38 \, \text{moles} \][/tex]
2. Calculate the volume of HCl gas at S.T.P:
At standard temperature and pressure (S.T.P), 1 mole of any ideal gas occupies [tex]\(22.4 \, \text{dm}^3\)[/tex] of volume.
To find the volume of [tex]\(7.38\)[/tex] moles of HCl gas, we use the relationship:
[tex]\[ \text{Volume of HCl gas at S.T.P} = \text{Number of moles of HCl} \times \text{Volume of one mole of gas at S.T.P} \][/tex]
Given:
- Number of moles of HCl = [tex]\(7.38\)[/tex]
- Volume of one mole of gas at S.T.P = [tex]\(22.4 \, \text{dm}^3\)[/tex]
[tex]\[ \text{Volume of HCl gas at S.T.P} = 7.38 \, \text{moles} \times 22.4 \, \text{dm}^3/\text{mole} = 165.312 \, \text{dm}^3 \][/tex]
Therefore, the volume of hydrogen chloride gas at S.T.P that would yield a [tex]\(1.2 \, \text{dm}^3\)[/tex] solution with a molarity of [tex]\(6.15 \, \text{M}\)[/tex] is [tex]\(165.312 \, \text{dm}^3\)[/tex].
