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Chapter 8: Problem 33

The number of \(\mathrm{F}^{-}\) ions in \(4.2 \mathrm{~g} \mathrm{AlF}_{3}\) is \((\mathrm{Al}=27, \mathrm{~F}=19)\) (a) \(0.05\) (b) \(9.03 \times 10^{22}\) (c) \(3.01 \times 10^{22}\) (d) \(0.15\)

Short Answer

Expert verified
Answer: Approximately 9.03 × 10^22 F- ions.

Step by step solution

01

Find the number of moles of AlF3

To find the number of moles of AlF3, we need to know the molar mass of AlF3. We are given the molar mass of Al and F in the problem statement: Molar mass of Al = 27 g/mol Molar mass of F = 19 g/mol Thus, the molar mass of AlF3 = (1 × 27) + (3 × 19) = 27 + 57 = 84 g/mol Now we can find the number of moles of AlF3 by dividing the given mass (4.2 g) by the molar mass (84 g/mol): Number of moles of AlF3 = 4.2 g / 84 g/mol = 0.05 mol.
02

Find the mole ratio between AlF3 and F- ions

We need to find the mole ratio between AlF3 and F- ions in order to find the number of moles of F- ions present. In one formula unit of AlF3, there is 1 Al atom and 3 F atoms. Since there are 3 F- ions for every AlF3 unit, the mole ratio between AlF3 and F- ions is 1:3.
03

Calculate the number of moles of F- ions

Now that we know the mole ratio between AlF3 and F- ions, we can find the number of moles of F- ions. Number of moles of F- ions = (Number of moles of AlF3) × (Mole ratio between AlF3 and F- ions) Number of moles of F- ions = 0.05 mol × (3 moles of F- ions / 1 mole of AlF3) = 0.15 mol.
04

Convert moles of F- ions to the number of ions using Avogadro's number

Now we will convert the moles of F- ions to the actual number of ions using Avogadro's number (6.022 x 10^23 ions/mol): Number of F- ions = (Number of moles of F- ions) × (Avogadro's number) Number of F- ions = 0.15 mol × (6.022 × 10^23 ions/mol) = 9.03 × 10^22 ions Thus, the number of F- ions in 4.2 g of AlF3 is approximately 9.03 × 10^22 ions. The correct answer is (b).

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Most popular questions from this chapter

A compound having the empirical formula, \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}\), has a molecular weight of \(170 \pm 5 .\) The molecular formula of the compound is (a) \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{2}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{3}\) (d) \(\mathrm{C}_{9} \mathrm{H}_{12} \mathrm{O}_{3}\)

A volume of \(10 \mathrm{ml}\) of an oxide of nitrogen was taken in a eudiometer tube and mixed with hydrogen until the volume was \(28 \mathrm{ml} .\) On sparking, the resulting mixture occupied \(18 \mathrm{ml}\). To this mixture, oxygen was added when the volume came to \(27 \mathrm{ml}\) and on explosion again, the volume fall to \(15 \mathrm{ml}\). Find the molecular weight of the oxide of nitrogen originally taken in eudiometer tube. All measurements were made at STP.

How many grams of \(90 \%\) pure \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) can be produced from \(250 \mathrm{~g}\) of \(95 \%\) pure \(\mathrm{NaCl} ?\) (a) \(640.6 \mathrm{~g}\) (b) \(288.2 \mathrm{~g}\) (c) \(259.4 \mathrm{~g}\) (d) \(320.3 \mathrm{~g}\)

An amount of \(0.3\) mole of \(\mathrm{SrCl}_{2}\) is mixed with \(0.2\) mole of \(\mathrm{K}_{3} \mathrm{PO}_{4}\). The maximum moles of \(\mathrm{KCl}\) which may form is (a) \(0.6\) (b) \(0.5\) (c) \(0.3\) (d) \(0.1\)

What mass of carbon disulphide, \(\mathrm{CS}_{2}\), can be completely oxidized to \(\mathrm{SO}_{2}\) and \(\mathrm{CO}_{2}\) by the oxygen liberated when \(325 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{O}_{2}\) react with water? (a) \(316.67 \mathrm{~g}\) (b) \(52.78 \mathrm{~g}\) (c) \(633.33 \mathrm{~g}\) (d) \(211.11 \mathrm{~g}\)
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