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Answer :
To find out how many grams of Cu(NO₃)₂ are required to make a 1.00 L of 0.420 M solution, follow these steps:
1. Determine the Molar Mass of Cu(NO₃)₂:
- Copper (Cu): 63.55 g/mol
- Nitrogen (N): 14.01 g/mol
- Oxygen (O): 16.00 g/mol
Since Cu(NO₃)₂ has one copper atom, two nitrogen atoms, and six oxygen atoms, its molar mass is calculated as follows:
[tex]\[
\text{Molar Mass of Cu(NO}_3\text{)}_2 = 63.55 + 2(14.01 + 3 \times 16.00)
\][/tex]
[tex]\[
= 63.55 + 2(14.01 + 48.00)
\][/tex]
[tex]\[
= 63.55 + 2(62.01)
\][/tex]
[tex]\[
= 63.55 + 124.02 = 187.57 \text{ g/mol}
\][/tex]
2. Calculate the Amount in Grams Needed:
- Molarity (M) is defined as moles of solute per liter of solution. Given a 0.420 M solution and a volume of 1.00 L, the moles of Cu(NO₃)₂ needed is:
[tex]\[
\text{Moles of Cu(NO}_3\text{)}_2 = 0.420 \text{ moles/L} \times 1.00 \text{ L} = 0.420 \text{ moles}
\][/tex]
- To find the grams required, multiply the moles by the molar mass:
[tex]\[
\text{Grams of Cu(NO}_3\text{)}_2 = 0.420 \text{ moles} \times 187.57 \text{ g/mol} = 78.7784 \text{ grams}
\][/tex]
Therefore, approximately 78.78 grams of Cu(NO₃)₂ are required. However, since this answer is not among the given options, it seems like there's a mistake either in the options or the assumptions. But your calculated result should be considered correct as per the steps above.
1. Determine the Molar Mass of Cu(NO₃)₂:
- Copper (Cu): 63.55 g/mol
- Nitrogen (N): 14.01 g/mol
- Oxygen (O): 16.00 g/mol
Since Cu(NO₃)₂ has one copper atom, two nitrogen atoms, and six oxygen atoms, its molar mass is calculated as follows:
[tex]\[
\text{Molar Mass of Cu(NO}_3\text{)}_2 = 63.55 + 2(14.01 + 3 \times 16.00)
\][/tex]
[tex]\[
= 63.55 + 2(14.01 + 48.00)
\][/tex]
[tex]\[
= 63.55 + 2(62.01)
\][/tex]
[tex]\[
= 63.55 + 124.02 = 187.57 \text{ g/mol}
\][/tex]
2. Calculate the Amount in Grams Needed:
- Molarity (M) is defined as moles of solute per liter of solution. Given a 0.420 M solution and a volume of 1.00 L, the moles of Cu(NO₃)₂ needed is:
[tex]\[
\text{Moles of Cu(NO}_3\text{)}_2 = 0.420 \text{ moles/L} \times 1.00 \text{ L} = 0.420 \text{ moles}
\][/tex]
- To find the grams required, multiply the moles by the molar mass:
[tex]\[
\text{Grams of Cu(NO}_3\text{)}_2 = 0.420 \text{ moles} \times 187.57 \text{ g/mol} = 78.7784 \text{ grams}
\][/tex]
Therefore, approximately 78.78 grams of Cu(NO₃)₂ are required. However, since this answer is not among the given options, it seems like there's a mistake either in the options or the assumptions. But your calculated result should be considered correct as per the steps above.
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