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QuestionWhich expression calculates the number of particles in 2.0 g of He? Options:
1. (2.0 g4.0 g/mol)6.02×1023 particles 1 mol\left(\frac{2.0 \mathrm{~g}}{4.0 \mathrm{~g} / \mathrm{mol}}\right) \frac{6.02 \times 10^{23} \text { particles }}{1 \mathrm{~mol}}
2. 6.02×1023 particles /mol4.0 g/mol\frac{6.02 \times 10^{23} \text { particles } / \mathrm{mol}}{4.0 \mathrm{~g} / \mathrm{mol}}
3. (4.0 g/mol2.0 g)6.02×1023particles1 mol\left(\frac{4.0 \mathrm{~g} / \mathrm{mol}}{2.0 \mathrm{~g}}\right) \frac{6.02 \times 10^{23} \mathrm{particles}}{1 \mathrm{~mol}}
4. 6.02×1023particles/mol2.0 g\frac{6.02 \times 10^{23} \mathrm{particles} / \mathrm{mol}}{2.0 \mathrm{~g}}

Studdy Solution

STEP 1

Assumptions1. The mass of helium (He) given is.0 g. The molar mass of helium (He) is4.0 g/mol3. Avogadro's number is 6.02×10236.02 \times10^{23} particles/mol4. The number of particles in a given mass of a substance can be calculated using the formula Number of particles=(Given massMolar mass)×Avogadro’s number\text{Number of particles} = \left(\frac{\text{Given mass}}{\text{Molar mass}}\right) \times \text{Avogadro's number}

STEP 2

We need to find the expression that correctly applies the formula from the assumptions. The correct expression should first divide the given mass by the molar mass, and then multiply the result by Avogadro's number.

STEP 3

Looking at the first expression, we see that it correctly divides the given mass (2.0 g) by the molar mass (.0 g/mol), and then multiplies the result by Avogadro's number (6.02×10236.02 \times10^{23} particles/mol).
(2.0 g.0 g/mol)6.02×1023 particles 1 mol\left(\frac{2.0 \mathrm{~g}}{.0 \mathrm{~g} / \mathrm{mol}}\right) \frac{6.02 \times10^{23} \text { particles }}{1 \mathrm{~mol}}

STEP 4

The second expression incorrectly divides Avogadro's number by the molar mass, instead of multiplying the result of the division of the given mass by the molar mass by Avogadro's number.
6.02×1023 particles /mol4.0 g/mol\frac{6.02 \times10^{23} \text { particles } / \mathrm{mol}}{4.0 \mathrm{~g} / \mathrm{mol}}

STEP 5

The third expression incorrectly divides the molar mass by the given mass, instead of dividing the given mass by the molar mass. It then correctly multiplies the result by Avogadro's number.
(4.0 g/mol2.0 g).02×1023particles1 mol\left(\frac{4.0 \mathrm{~g} / \mathrm{mol}}{2.0 \mathrm{~g}}\right) \frac{.02 \times10^{23} \mathrm{particles}}{1 \mathrm{~mol}}

STEP 6

The fourth expression incorrectly divides Avogadro's number by the given mass, instead of multiplying the result of the division of the given mass by the molar mass by Avogadro's number.
6.02×1023particles/mol2.0 g\frac{6.02 \times10^{23} \mathrm{particles} / \mathrm{mol}}{2.0 \mathrm{~g}}

STEP 7

Based on the analysis in steps3-6, the first expression is the correct one that gives the number of particles in2.0 g of He.
So, the correct numerical expression is(2.0 g4.0 g/mol)6.02×1023 particles 1 mol\left(\frac{2.0 \mathrm{~g}}{4.0 \mathrm{~g} / \mathrm{mol}}\right) \frac{6.02 \times10^{23} \text { particles }}{1 \mathrm{~mol}}

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