Question

use the molar mass of al₂(so₄)₃ to calculate the number of formula units in 10.0 g of as. everything about a chemical formula is a direct proportion. we calculated the molar mass of aluminum sulfate as 342.15 g/mol. in the previous problem we calculated that there are 0.0292274 moles in 10.0 g. now we need to find out how many formula units there are in 10.0 g. a formula unit is one molecule of an ionic compound. the conversion factor is 1 mol = 6.022 x 10²³ things, where things can be atoms, molecules, or formula units. (_mol as)(_______) = _fu a. 0 b. 1 c. 2 d. 3 e. 4 f. 342.15 g. 10.0 h. 56.1 i. 0.0292274 j. 3420 k. 6.022 x 10²³ l. 1.76 x 10²² m. 4.85 x 10⁻²⁶ n. 1.76 x 10⁻²⁴

Explanation:

Step1: Determine the number of moles

We are given that the number of moles of \(Al_2(SO_4)_3\) (AS) is \(n = 0.0292274\) mol (from previous calculation).

Step2: Use Avogadro's number

Avogadro's number \(N_A=6.022\times 10^{23}\) formula units (FU)/mol. The conversion factor is \(\frac{6.022\times 10^{23}\text{ FU}}{1\text{ mol AS}}\).
We use the formula \(N = n\times N_A\), where \(n\) is the number of moles and \(N_A\) is Avogadro's number.
Substituting \(n = 0.0292274\) mol and \(N_A=6.022\times 10^{23}\text{ FU/mol}\) into the formula:
\(N=(0.0292274\text{ mol AS})\times\frac{6.022\times 10^{23}\text{ FU}}{1\text{ mol AS}}\)

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Answer:

1. \(K\) (since \(1\) mol has \(6.022\times 10^{23}\) formula units)
2. \(I\) (the number of moles of \(Al_2(SO_4)_3\) is \(0.0292274\) mol)
3. \(L\) (\(1.76\times 10^{22}\) formula units)
4. \(B\) (the coefficient for moles in the conversion factor is \(1\))