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Sagot :
To determine the mass of the solution, let's break the problem down into clear steps, using the given data:
1. Determine the volume and density of hydrochloric acid (HCl):
- Volume of HCl: \( 40.0 \) mL
- Density of HCl: \( 1.02 \) g/mL
2. Calculate the mass of hydrochloric acid (HCl):
- Mass of HCl can be calculated by multiplying its volume by its density.
[tex]\[ \text{Mass}_{\text{HCl}} = \text{Volume}_{\text{HCl}} \times \text{Density}_{\text{HCl}} \][/tex]
Substituting the given values:
[tex]\[ \text{Mass}_{\text{HCl}} = 40.0 \text{ mL} \times 1.02 \text{ g/mL} = 40.8 \text{ g} \][/tex]
3. Given the mass of magnesium (Mg) added:
- Mass of Mg: \( 0.486 \) g
4. Calculate the total mass of the solution:
- The total mass of the solution is the sum of the mass of HCl and the mass of Mg.
[tex]\[ \text{Mass}_{\text{solution}} = \text{Mass}_{\text{HCl}} + \text{Mass}_{\text{Mg}} \][/tex]
Substituting the values we have:
[tex]\[ \text{Mass}_{\text{solution}} = 40.8 \text{ g} + 0.486 \text{ g} = 41.286 \text{ g} \][/tex]
Thus, the mass of the solution is [tex]\( 41.286 \)[/tex] g.
1. Determine the volume and density of hydrochloric acid (HCl):
- Volume of HCl: \( 40.0 \) mL
- Density of HCl: \( 1.02 \) g/mL
2. Calculate the mass of hydrochloric acid (HCl):
- Mass of HCl can be calculated by multiplying its volume by its density.
[tex]\[ \text{Mass}_{\text{HCl}} = \text{Volume}_{\text{HCl}} \times \text{Density}_{\text{HCl}} \][/tex]
Substituting the given values:
[tex]\[ \text{Mass}_{\text{HCl}} = 40.0 \text{ mL} \times 1.02 \text{ g/mL} = 40.8 \text{ g} \][/tex]
3. Given the mass of magnesium (Mg) added:
- Mass of Mg: \( 0.486 \) g
4. Calculate the total mass of the solution:
- The total mass of the solution is the sum of the mass of HCl and the mass of Mg.
[tex]\[ \text{Mass}_{\text{solution}} = \text{Mass}_{\text{HCl}} + \text{Mass}_{\text{Mg}} \][/tex]
Substituting the values we have:
[tex]\[ \text{Mass}_{\text{solution}} = 40.8 \text{ g} + 0.486 \text{ g} = 41.286 \text{ g} \][/tex]
Thus, the mass of the solution is [tex]\( 41.286 \)[/tex] g.
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