Respuesta :
Answer:
- 2.0 X 10²² molecules of AgNO₃
Explanation:
Given 5.50 grams of AgNO₃, how many formula units of AgNO₃ are contained in the 5.50 grams of AgNO₃? (formula wt. of AgNO₃ = 169,87 g/mole).
Solution:
- 1st convert grams to moles => moles AgNO₃ = 5.50 g AgNO₃ / 169.841g AgNO₃ = 0.032377 mole AgNO₃
- 2nd calculate number of particles in 0.032377 mole AgNO₃ . Number AgNO₃ molecules in 0.032377 mole AgNO₃ = 0.032377 mole AgNO₃ X 6.023 x 10²³ molecules of AgNO₃ /1.0 mole AgNO₃ = 1.95007 X 10²² molecules of AgNO₃ in 5.50 grams of AgNO₃.
- The appropriate form of the answer should contain 2 Sig,Figs. based on the data point having the least number of sig.figs. in the given data. This then is 5.50 grams of AgNO₃ which has 2 sig.figs.
- Therefore, the number of AgNO₃ molecules in 5.5 grams of AgNO₃ = 2.0 X 10²² molecules of AgNO₃. (Note: 1.95 rounds to 2.0).
The number of molecules of silver nitrate in 5.5 grams has been [tex]1.93\;\times\;10^2^2[/tex]. Thus, option A is correct.
The formula unit has been given as the number of molecules of the compound. The number of molecules in a mole of compound has been given by the Avogadro law.
According to the Avogadro law, the number of molecules in a mole of sample has been equivalent to the Avogadro number, i.e. [tex]6.023\;\times\;10^2^3[/tex].
Computation for formula unit of Silver nitrate
The given mass of silver nitrate has been 5.50 grams.
The molar mass of silver nitrate has been 169.84 g/mol.
The moles of silver nitrate has been:
[tex]\rm Moles=\dfrac{Mass}{Molar\;mass} \\\\
Moles=\dfrac{5.5}{169.84}\;mol\\\\
Moles=0.032\;mol[/tex]
The moles of silver nitrate in the sample has been 0.032 mol.
The number of molecules of silver nitrate has been:
[tex]\rm 1\;mole=6.023\;\times\;10^2^3\;molecules\\ 0.032\;mol=0.032\;\times\;6.023\;\times\;10^2^3\;molecules\\ 0.032\;mol=1.93\;\times\;10^2^2\;molecules[/tex]
The number of molecules of silver nitrate in 5.5 grams has been [tex]1.93\;\times\;10^2^2[/tex]. Thus, option A is correct.
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