10.2 conversions between moles and atoms\n1.) what is the mass of 5.119 x 10^23 molecules of copper sulfate…

10.2 conversions between moles and atoms\n1.) what is the mass of 5.119 x 10^23 molecules of copper sulfate (cuso₄)? reminder to use dimensional analysis from unit 4 to work out the problems. you will have to work them out this way on the test.\n(1 point)\n○ 159.6 g\n○ 94.88 g\n○ 0.85 g\n○ 135.7 g\n2.) how many moles of iron in a sample that contains 7.91 x 10^23 atoms of iron? (1 point)\n○ 7.91 mol fe\n○ 0.761 mol fe\n○ 1.31 mol fe\n○ 47.6 mol fe\n3.) how many moles would 1.00 x 10^23 molecules of methane (ch₄) represent? (1 point)\n○ 16.6 mol\n○ 0.166 mol\n○ 1.66 mol
Answer
Explanation:
Step1: Recall Avogadro's number
Avogadro's number ($N_A$) is $6.022\times 10^{23}$ molecules/mol.
Step2: Calculate moles for each problem
Problem 1
First, find the number of moles of $CuSO_4$. The number of moles $n=\frac{5.119\times 10^{23}\text{ molecules}}{6.022\times 10^{23}\text{ molecules/mol}}\approx0.85\text{ mol}$. The molar - mass of $CuSO_4$: $M_{CuSO_4}=63.55 + 32.07+4\times16.00 = 159.62\text{ g/mol}$. Then the mass $m = n\times M=0.85\text{ mol}\times159.62\text{ g/mol}\approx135.7\text{ g}$.
Problem 2
The number of moles of Fe, $n=\frac{7.91\times 10^{23}\text{ atoms}}{6.022\times 10^{23}\text{ atoms/mol}}\approx1.31\text{ mol}$.
Problem 3
The number of moles of $CH_4$, $n=\frac{1.00\times 10^{23}\text{ molecules}}{6.022\times 10^{23}\text{ molecules/mol}}\approx0.166\text{ mol}$.
Answer:
- D. 135.7 g
- C. 1.31 mol Fe
- B. 0.166 mol