Answer :
Certainly! Let's go through each part step by step to determine the molarity of the solute in each solution.
### Part a: [tex]$8.33 \, \text{g}\, \text{MgNH}_4 \text{PO}_4$[/tex] in [tex]$600 \, \text{mL}$[/tex] Solution
1. Calculate the molar mass of [tex]$\text{MgNH}_4\text{PO}_4$[/tex]:
[tex]\[ \text{Molar mass of MgNH}_4\text{PO}_4 = 137.33 \, (\text{Mg}) + 1.008 \, (\text{H}) + 14.01 \, (\text{N}) + 4 \times 15.999 \, (\text{O}) \][/tex]
[tex]\[ \text{Molar mass of MgNH}_4\text{PO}_4 = 137.33 + 1.008 + 14.01 + 63.996 = 216.344 \, \text{g/mol} \][/tex]
2. Convert the volume from milliliters to liters:
[tex]\[ 600 \, \text{mL} = \frac{600}{1000} = 0.6 \, \text{L} \][/tex]
3. Calculate the number of moles of [tex]$\text{MgNH}_4\text{PO}_4$[/tex]:
[tex]\[ \text{Moles of MgNH}_4\text{PO}_4 = \frac{8.33 \, \text{g}}{216.344 \, \text{g/mol}} = 0.038503 \, \text{mol} \][/tex]
4. Calculate the molarity:
[tex]\[ \text{Molarity} = \frac{0.038503 \, \text{mol}}{0.6 \, \text{L}} = 0.064172 \, \text{mol/L} \][/tex]
Molarity = 0.064172 mol/L
### Part b: [tex]$10.0 \, \text{g} \, \text{NaCH}_3\text{COO}$[/tex] in [tex]$300 \, \text{mL}$[/tex] Solution
1. Calculate the molar mass of [tex]$\text{NaCH}_3\text{COO}$[/tex]:
[tex]\[ \text{Molar mass of NaCH}_3\text{COO} = 22.99 \, (\text{Na}) + 12.01 \, (\text{C}) + 3 \times 1.008 \, (\text{H}) + 2 \times 15.999 \, (\text{O}) \][/tex]
[tex]\[ \text{Molar mass of NaCH}_3\text{COO} = 22.99 + 12.01 + 3.024 + 31.998 = 70.022 \, \text{g/mol} \][/tex]
2. Convert the volume from milliliters to liters:
[tex]\[ 300 \, \text{mL} = \frac{300}{1000} = 0.3 \, \text{L} \][/tex]
3. Calculate the number of moles of [tex]$\text{NaCH}_3\text{COO}$[/tex]:
[tex]\[ \text{Moles of NaCH}_3\text{COO} = \frac{10.0 \, \text{g}}{70.022 \, \text{g/mol}} = 0.142812 \, \text{mol} \][/tex]
4. Calculate the molarity:
[tex]\[ \text{Molarity} = \frac{0.142812 \, \text{mol}}{0.3 \, \text{L}} = 0.476041 \, \text{mol/L} \][/tex]
Molarity = 0.476041 mol/L
### Part c: [tex]$8.24 \, \text{g} \, \text{CaC}_2\text{O}_4$[/tex] in [tex]$750 \, \text{mL}$[/tex] Solution
1. Calculate the molar mass of [tex]$\text{CaC}_2\text{O}_4$[/tex]:
[tex]\[ \text{Molar mass of CaC}_2\text{O}_4 = 40.08 \, (\text{Ca}) + 2 \times 12.01 \, (\text{C}) + 4 \times 15.999 \, (\text{O}) \][/tex]
[tex]\[ \text{Molar mass of CaC}_2\text{O}_4 = 40.08 + 24.02 + 63.996 = 128.096 \, \text{g/mol} \][/tex]
2. Convert the volume from milliliters to liters:
[tex]\[ 750 \, \text{mL} = \frac{750}{1000} = 0.75 \, \text{L} \][/tex]
3. Calculate the number of moles of [tex]$\text{CaC}_2\text{O}_4$[/tex]:
[tex]\[ \text{Moles of CaC}_2\text{O}_4 = \frac{8.24 \, \text{g}}{128.096 \, \text{g/mol}} = 0.064327 \, \text{mol} \][/tex]
4. Calculate the molarity:
[tex]\[ \text{Molarity} = \frac{0.064327 \, \text{mol}}{0.75 \, \text{L}} = 0.085769 \, \text{mol/L} \][/tex]
Molarity = 0.085769 mol/L
### Part d: [tex]$2.03 \, \text{g} \left(\text{NH}_4\right)_2\text{SO}_4$[/tex] in [tex]$400 \, \text{mL}$[/tex] Solution
1. Calculate the molar mass of [tex]$\left(\text{NH}_4\right)_2\text{SO}_4$[/tex]:
[tex]\[ \text{Molar mass of } \left(\text{NH}_4\right)_2\text{SO}_4 = 2 \times (1.008 + 4 \times 1.008) + 32.06 \, (\text{S}) + 4 \times 15.999 \, (\text{O}) \][/tex]
[tex]\[ \text{Molar mass of } \left(\text{NH}_4\right)_2\text{SO}_4 = 2 \times 5.04 + 32.06 + 63.996 = 132.14 \, \text{g/mol} \][/tex]
2. Convert the volume from milliliters to liters:
[tex]\[ 400 \, \text{mL} = \frac{400}{1000} = 0.4 \, \text{L} \][/tex]
3. Calculate the number of moles of [tex]$\left(\text{NH}_4\right)_2\text{SO}_4$[/tex]:
[tex]\[ \text{Moles of } \left(\text{NH}_4\right)_2\text{SO}_4 = \frac{2.03 \, \text{g}}{132.14 \, \text{g/mol}} = 0.019126 \, \text{mol} \][/tex]
4. Calculate the molarity:
[tex]\[ \text{Molarity} = \frac{0.019126 \, \text{mol}}{0.4 \, \text{L}} = 0.047816 \, \text{mol/L} \][/tex]
Molarity = 0.047816 mol/L
Summary of Molarities:
a. Molarity of [tex]$8.33 \, \text{g} \, \text{MgNH}_4\text{PO}_4$[/tex] in [tex]$600 \, \text{mL}$[/tex] solution = [tex]$0.064172 \, \text{mol/L}$[/tex]
b. Molarity of [tex]$10.0 \, \text{g} \, \text{NaCH}_3\text{COO}$[/tex] in [tex]$300 \, \text{mL}$[/tex] solution = [tex]$0.476041 \, \text{mol/L}$[/tex]
c. Molarity of [tex]$8.24 \, \text{g} \, \text{CaC}_2\text{O}_4$[/tex] in [tex]$750 \, \text{mL}$[/tex] solution = [tex]$0.085769 \, \text{mol/L}$[/tex]
d. Molarity of [tex]$2.03 \, \text{g} \, \left(\text{NH}_4\right)_2\text{SO}_4$[/tex] in [tex]$400 \, \text{mL}$[/tex] solution = [tex]$0.047816 \, \text{mol/L}$[/tex]
### Part a: [tex]$8.33 \, \text{g}\, \text{MgNH}_4 \text{PO}_4$[/tex] in [tex]$600 \, \text{mL}$[/tex] Solution
1. Calculate the molar mass of [tex]$\text{MgNH}_4\text{PO}_4$[/tex]:
[tex]\[ \text{Molar mass of MgNH}_4\text{PO}_4 = 137.33 \, (\text{Mg}) + 1.008 \, (\text{H}) + 14.01 \, (\text{N}) + 4 \times 15.999 \, (\text{O}) \][/tex]
[tex]\[ \text{Molar mass of MgNH}_4\text{PO}_4 = 137.33 + 1.008 + 14.01 + 63.996 = 216.344 \, \text{g/mol} \][/tex]
2. Convert the volume from milliliters to liters:
[tex]\[ 600 \, \text{mL} = \frac{600}{1000} = 0.6 \, \text{L} \][/tex]
3. Calculate the number of moles of [tex]$\text{MgNH}_4\text{PO}_4$[/tex]:
[tex]\[ \text{Moles of MgNH}_4\text{PO}_4 = \frac{8.33 \, \text{g}}{216.344 \, \text{g/mol}} = 0.038503 \, \text{mol} \][/tex]
4. Calculate the molarity:
[tex]\[ \text{Molarity} = \frac{0.038503 \, \text{mol}}{0.6 \, \text{L}} = 0.064172 \, \text{mol/L} \][/tex]
Molarity = 0.064172 mol/L
### Part b: [tex]$10.0 \, \text{g} \, \text{NaCH}_3\text{COO}$[/tex] in [tex]$300 \, \text{mL}$[/tex] Solution
1. Calculate the molar mass of [tex]$\text{NaCH}_3\text{COO}$[/tex]:
[tex]\[ \text{Molar mass of NaCH}_3\text{COO} = 22.99 \, (\text{Na}) + 12.01 \, (\text{C}) + 3 \times 1.008 \, (\text{H}) + 2 \times 15.999 \, (\text{O}) \][/tex]
[tex]\[ \text{Molar mass of NaCH}_3\text{COO} = 22.99 + 12.01 + 3.024 + 31.998 = 70.022 \, \text{g/mol} \][/tex]
2. Convert the volume from milliliters to liters:
[tex]\[ 300 \, \text{mL} = \frac{300}{1000} = 0.3 \, \text{L} \][/tex]
3. Calculate the number of moles of [tex]$\text{NaCH}_3\text{COO}$[/tex]:
[tex]\[ \text{Moles of NaCH}_3\text{COO} = \frac{10.0 \, \text{g}}{70.022 \, \text{g/mol}} = 0.142812 \, \text{mol} \][/tex]
4. Calculate the molarity:
[tex]\[ \text{Molarity} = \frac{0.142812 \, \text{mol}}{0.3 \, \text{L}} = 0.476041 \, \text{mol/L} \][/tex]
Molarity = 0.476041 mol/L
### Part c: [tex]$8.24 \, \text{g} \, \text{CaC}_2\text{O}_4$[/tex] in [tex]$750 \, \text{mL}$[/tex] Solution
1. Calculate the molar mass of [tex]$\text{CaC}_2\text{O}_4$[/tex]:
[tex]\[ \text{Molar mass of CaC}_2\text{O}_4 = 40.08 \, (\text{Ca}) + 2 \times 12.01 \, (\text{C}) + 4 \times 15.999 \, (\text{O}) \][/tex]
[tex]\[ \text{Molar mass of CaC}_2\text{O}_4 = 40.08 + 24.02 + 63.996 = 128.096 \, \text{g/mol} \][/tex]
2. Convert the volume from milliliters to liters:
[tex]\[ 750 \, \text{mL} = \frac{750}{1000} = 0.75 \, \text{L} \][/tex]
3. Calculate the number of moles of [tex]$\text{CaC}_2\text{O}_4$[/tex]:
[tex]\[ \text{Moles of CaC}_2\text{O}_4 = \frac{8.24 \, \text{g}}{128.096 \, \text{g/mol}} = 0.064327 \, \text{mol} \][/tex]
4. Calculate the molarity:
[tex]\[ \text{Molarity} = \frac{0.064327 \, \text{mol}}{0.75 \, \text{L}} = 0.085769 \, \text{mol/L} \][/tex]
Molarity = 0.085769 mol/L
### Part d: [tex]$2.03 \, \text{g} \left(\text{NH}_4\right)_2\text{SO}_4$[/tex] in [tex]$400 \, \text{mL}$[/tex] Solution
1. Calculate the molar mass of [tex]$\left(\text{NH}_4\right)_2\text{SO}_4$[/tex]:
[tex]\[ \text{Molar mass of } \left(\text{NH}_4\right)_2\text{SO}_4 = 2 \times (1.008 + 4 \times 1.008) + 32.06 \, (\text{S}) + 4 \times 15.999 \, (\text{O}) \][/tex]
[tex]\[ \text{Molar mass of } \left(\text{NH}_4\right)_2\text{SO}_4 = 2 \times 5.04 + 32.06 + 63.996 = 132.14 \, \text{g/mol} \][/tex]
2. Convert the volume from milliliters to liters:
[tex]\[ 400 \, \text{mL} = \frac{400}{1000} = 0.4 \, \text{L} \][/tex]
3. Calculate the number of moles of [tex]$\left(\text{NH}_4\right)_2\text{SO}_4$[/tex]:
[tex]\[ \text{Moles of } \left(\text{NH}_4\right)_2\text{SO}_4 = \frac{2.03 \, \text{g}}{132.14 \, \text{g/mol}} = 0.019126 \, \text{mol} \][/tex]
4. Calculate the molarity:
[tex]\[ \text{Molarity} = \frac{0.019126 \, \text{mol}}{0.4 \, \text{L}} = 0.047816 \, \text{mol/L} \][/tex]
Molarity = 0.047816 mol/L
Summary of Molarities:
a. Molarity of [tex]$8.33 \, \text{g} \, \text{MgNH}_4\text{PO}_4$[/tex] in [tex]$600 \, \text{mL}$[/tex] solution = [tex]$0.064172 \, \text{mol/L}$[/tex]
b. Molarity of [tex]$10.0 \, \text{g} \, \text{NaCH}_3\text{COO}$[/tex] in [tex]$300 \, \text{mL}$[/tex] solution = [tex]$0.476041 \, \text{mol/L}$[/tex]
c. Molarity of [tex]$8.24 \, \text{g} \, \text{CaC}_2\text{O}_4$[/tex] in [tex]$750 \, \text{mL}$[/tex] solution = [tex]$0.085769 \, \text{mol/L}$[/tex]
d. Molarity of [tex]$2.03 \, \text{g} \, \left(\text{NH}_4\right)_2\text{SO}_4$[/tex] in [tex]$400 \, \text{mL}$[/tex] solution = [tex]$0.047816 \, \text{mol/L}$[/tex]
