Answer :
Sure, let's go through the problems step by step.
### Question 4:
How many moles of NaCl are present in 600 mL of a 1.55 M NaCl solution?
1. Convert volume from mL to liters:
Since molarity (M) is given in moles per liter, we need to convert the volume from milliliters to liters.
[tex]\[
600 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.600 \, \text{L}
\][/tex]
2. Use the molarity formula:
Molarity (M) is defined as moles of solute per liter of solution.
[tex]\[
\text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}}
\][/tex]
3. Calculate the moles of NaCl:
Rearrange the formula to solve for moles of NaCl:
[tex]\[
\text{moles of NaCl} = \text{Molarity} \times \text{liters of solution}
\][/tex]
[tex]\[
\text{moles of NaCl} = 1.55 \, \text{M} \times 0.600 \, \text{L} = 0.93 \, \text{moles}
\][/tex]
Therefore, there are 0.93 moles of NaCl in 600 mL of a 1.55 M NaCl solution.
### Question 5:
How many moles of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] are present in 1.63 liters of a 0.954 M solution?
1. Given values:
- Volume of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] solution: 1.63 liters
- Molarity of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] solution: 0.954 M
2. Use the molarity formula:
As before, we use the molarity formula:
[tex]\[
\text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}}
\][/tex]
3. Calculate the moles of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]:
Rearrange the formula to solve for moles of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]:
[tex]\[
\text{moles of H}_2\text{SO}_4 = \text{Molarity} \times \text{liters of solution}
\][/tex]
[tex]\[
\text{moles of H}_2\text{SO}_4 = 0.954 \, \text{M} \times 1.63 \, \text{L} = 1.555 \, \text{moles}
\][/tex]
Therefore, there are 1.555 moles of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] in 1.63 liters of a 0.954 M H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] solution.
To summarize:
- In 600 mL of a 1.55 M NaCl solution, there are 0.93 moles of NaCl.
- In 1.63 liters of a 0.954 M H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] solution, there are 1.555 moles of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex].
### Question 4:
How many moles of NaCl are present in 600 mL of a 1.55 M NaCl solution?
1. Convert volume from mL to liters:
Since molarity (M) is given in moles per liter, we need to convert the volume from milliliters to liters.
[tex]\[
600 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.600 \, \text{L}
\][/tex]
2. Use the molarity formula:
Molarity (M) is defined as moles of solute per liter of solution.
[tex]\[
\text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}}
\][/tex]
3. Calculate the moles of NaCl:
Rearrange the formula to solve for moles of NaCl:
[tex]\[
\text{moles of NaCl} = \text{Molarity} \times \text{liters of solution}
\][/tex]
[tex]\[
\text{moles of NaCl} = 1.55 \, \text{M} \times 0.600 \, \text{L} = 0.93 \, \text{moles}
\][/tex]
Therefore, there are 0.93 moles of NaCl in 600 mL of a 1.55 M NaCl solution.
### Question 5:
How many moles of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] are present in 1.63 liters of a 0.954 M solution?
1. Given values:
- Volume of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] solution: 1.63 liters
- Molarity of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] solution: 0.954 M
2. Use the molarity formula:
As before, we use the molarity formula:
[tex]\[
\text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}}
\][/tex]
3. Calculate the moles of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]:
Rearrange the formula to solve for moles of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]:
[tex]\[
\text{moles of H}_2\text{SO}_4 = \text{Molarity} \times \text{liters of solution}
\][/tex]
[tex]\[
\text{moles of H}_2\text{SO}_4 = 0.954 \, \text{M} \times 1.63 \, \text{L} = 1.555 \, \text{moles}
\][/tex]
Therefore, there are 1.555 moles of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] in 1.63 liters of a 0.954 M H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] solution.
To summarize:
- In 600 mL of a 1.55 M NaCl solution, there are 0.93 moles of NaCl.
- In 1.63 liters of a 0.954 M H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] solution, there are 1.555 moles of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex].
