Question

1. you have been given a sample of aluminum measuring 30.00 cm by 10.00 cm by 1.00 cm. your teacher has asked you to determine the number of particles in the sample. you looked up the density of aluminum and found it to be 2.70 g/cm³.

Explanation:

Step 1: Calculate the volume of the aluminum sample

The sample is a rectangular prism, so the volume \( V \) is calculated by multiplying its length, width, and height. The dimensions are \( 30.00 \, \text{cm} \), \( 10.00 \, \text{cm} \), and \( 1.00 \, \text{cm} \). So, \( V = 30.00 \times 10.00 \times 1.00 \).
\( V = 300.0 \, \text{cm}^3 \)

Step 2: Calculate the mass of the aluminum sample

Density \( ho \) is given by the formula \( ho = \frac{m}{V} \), where \( m \) is mass and \( V \) is volume. We know the density of aluminum \( ho = 2.70 \, \text{g/cm}^3 \) and the volume \( V = 300.0 \, \text{cm}^3 \). Rearranging the formula to solve for mass, we get \( m = ho \times V \).
\( m = 2.70 \, \text{g/cm}^3 \times 300.0 \, \text{cm}^3 \)
\( m = 810 \, \text{g} \) (Note: We'll keep more precision for now, actually \( 2.70\times300.0 = 810. \, \text{g} \) with three significant figures from density? Wait, density is 2.70 (three sig figs), volume is 300.0 (four sig figs, but since it's a product, we go with three sig figs? Wait, 30.00×10.00×1.00: 30.00 has four, 10.00 has four, 1.00 has three. The least number of decimal places? No, for multiplication, it's the least number of significant figures. 1.00 has three, so volume is \( 30.00\times10.00\times1.00 = 300.0 \, \text{cm}^3 \) (but 1.00 has three, so maybe 3.00×10²? Wait, no, 30.00 is four, 10.00 is four, 1.00 is three. So the volume should be reported with three significant figures? Wait, 30.00×10.00 = 300.0 (four sig figs), then ×1.00 (three sig figs) gives 300. (three sig figs? Wait, 300.0×1.00 = 300. (the decimal is important here, but maybe the problem expects us to use the given numbers as is. Let's proceed with \( V = 300.0 \, \text{cm}^3 \) for now. Then mass \( m = 2.70 \times 300.0 = 810. \, \text{g} \) (three sig figs from density, so 810. g can be 8.10×10² g? Wait, no, 2.70 has three sig figs, 300.0 has four, so the product should have three sig figs. So 2.70×300.0 = 810. (but written as 8.10×10² g? Wait, maybe I'm overcomplicating. Let's just calculate the mass as 810 g for now, but actually, 2.70 g/cm³ × 300.0 cm³ = 810. g (three sig figs, so 8.10×10² g? Wait, 2.70 is three, 300.0 is four, so the result should have three sig figs. So 810. g is 8.10×10² g? Wait, no, 2.70×300.0 = 810. (the decimal after 810 indicates that the trailing zero is significant? Wait, maybe the volume is 30.00×10.00×1.00 = 300.0 cm³ (four sig figs), density is 2.70 (three sig figs), so mass is 2.70×300.0 = 810. g (three sig figs, so 8.10×10² g? Wait, 810. with a decimal is three sig figs? No, 810. has three sig figs? Wait, 810 without a decimal is ambiguous, but with a decimal, it's three (8,1,0). So 810. g is three sig figs.

Step 3: Calculate the number of moles of aluminum

The molar mass of aluminum (Al) is approximately \( 26.98 \, \text{g/mol} \). Moles \( n \) is calculated by \( n = \frac{m}{M} \), where \( m \) is mass and \( M \) is molar mass.
\( n = \frac{810. \, \text{g}}{26.98 \, \text{g/mol}} \)
\( n \approx 30.02 \, \text{mol} \) (let's calculate it more accurately: 810 ÷ 26.98 ≈ 30.02 mol)

Step 4: Calculate the number of particles (atoms) using Avogadro's number

Avogadro's number \( N_A = 6.022 \times 10^{23} \, \text{particles/mol} \). The number of particles \( N \) is given by \( N = n \times N_A \).
\( N = 30.02 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \)
\( N \approx 30.02 \times 6.022 \times 10^{23} \)
First, calculate 30.02×6.022: 30×6.022 = 180.66, 0.02×6.022 = 0.12044, so total ≈ 180.78044. Then multiply by \( 10^{23} \): \( N \approx 1.81 \times 10^{25} \) atoms? Wait, wait, 30.02×6.022×10²³: 30.02×6.022 = let's do it properly: 26.98 is molar mass, wait, I think I made a mistake in molar mass. Aluminum's molar mass is 26.98 g/mol. So mass is 810 g (wait, no, 2.70 g/cm³ × 300.0 cm³ = 810 g. Then moles = 810 g / 26.98 g/mol ≈ 30.02 mol. Then number of atoms = 30.02 mol × 6.022×10²³ atoms/mol ≈ 30.02×6.022×10²³. Let's calculate 30×6.022 = 180.66, 0.02×6.022 = 0.12044, so total 180.78044×10²³ = 1.8078044×10²⁵ ≈ 1.81×10²⁵ atoms. Wait, but let's check the significant figures. Density is 2.70 (three sig figs), volume is 30.00×10.00×1.00: 30.00 (four), 10.00 (four), 1.00 (three). So volume is 300.0 cm³ (but 1.00 has three, so volume is 3.00×10² cm³ (three sig figs). Then mass = 2.70 (three) × 3.00×10² (three) = 8.10×10² g (three sig figs). Then moles = 8.10×10² g / 26.98 g/mol ≈ 30.0 mol (three sig figs). Then number of atoms = 30.0 mol × 6.022×10²³ atoms/mol = 1.8066×10²⁵ ≈ 1.81×10²⁵ atoms (three sig figs).

Wait, maybe I messed up the volume calculation. Let's recalculate volume: 30.00 cm × 10.00 cm × 1.00 cm. 30.00×10.00 = 300.0, then ×1.00 = 300.0 cm³. So volume is 300.0 cm³ (four sig figs). Density is 2.70 g/cm³ (three sig figs). So mass = 2.70×300.0 = 810. g (three sig figs, because 2.70 has three). Then moles = 810. g / 26.98 g/mol ≈ 30.02 mol (four sig figs? No, 810. has three, 26.98 has four, so moles should have three: 30.0 mol). Then number of atoms = 30.0 mol × 6.022×10²³ atoms/mol = 1.8066×10²⁵ ≈ 1.81×10²⁵ atoms.

Answer:

The number of particles (atoms) in the aluminum sample is approximately \( \boldsymbol{1.81 \times 10^{25}} \) (or more precisely, following significant figures, if we consider the density as 2.70 (three), volume as 300.0 (four), but mass is 810. (three), so the final answer should have three significant figures, so \( 1.81 \times 10^{25} \) atoms.