Question

nₐ-avogadro constant
practice questions:

1. a sample contains 4.70×10⁻⁴ mol of gold, cu. how many atoms of copper are in the sample?(ans.2.83×10²⁰)
2. how many molecules are in a baloon that contains 0.275 mol of ammonia?(ans.1.66×10²³ molecules)

Explanation:

Step1: Recall Avogadro's Number

Avogadro's number (\(N_A\)) is \(6.022 \times 10^{23}\) particles (atoms, molecules, etc.) per mole. The formula to find the number of particles (\(N\)) is \(N = n \times N_A\), where \(n\) is the number of moles.

Step2: Solve for Question 1

For the copper sample, \(n = 4.70 \times 10^{-4}\) mol. Using \(N = n \times N_A\):
\(N = 4.70 \times 10^{-4}\ \text{mol} \times 6.022 \times 10^{23}\ \text{atoms/mol}\)
Calculate the product: \(4.70 \times 6.022 \approx 28.3034\), and \(10^{-4} \times 10^{23} = 10^{19}\), so \(28.3034 \times 10^{19} = 2.83 \times 10^{20}\) atoms (matches the given answer).

Step3: Solve for Question 2

For ammonia (\(NH_3\)), \(n = 0.275\) mol. Using \(N = n \times N_A\):
\(N = 0.275\ \text{mol} \times 6.022 \times 10^{23}\ \text{molecules/mol}\)
Calculate the product: \(0.275 \times 6.022 \approx 1.656\), so \(N \approx 1.66 \times 10^{23}\) molecules (matches the given answer).

Answer:

1. The number of copper atoms is \(\boldsymbol{2.83 \times 10^{20}}\).
2. The number of ammonia molecules is \(\boldsymbol{1.66 \times 10^{23}}\).