Question

Calculate the mass of $4.7 \times {10^{22}}$ molecules of lead iodide (II)
A. 36.1grams
B. 249.0grams
C. 108.3grams
D. 697.3grams

Answer 1

We can calculate the mass of lead iodide molecules by converting the molecules to moles using Avogadro number. After converting the molecules to moles, we have multiplied the obtained value with molar mass of lead iodide. We know that the value of Avogadro number is $6.022 \times {10^{23}}$.

Complete step by step answer:
Given data contains,
The molecules of lead (II) iodide are $4.7 \times {10^{22}}$ molecules.
Avogadro number is $6.022 \times {10^{23}}$.
We have to divide the molecules of lead (II) iodide by the Avogadro number to get the moles of lead (II) iodide. We can write the formula to calculate the moles of lead (II) iodide as,
${\text{Moles}} = \dfrac{{{\text{molecules}}}}{{6.022 \times {{10}^{23}}}}$
Let us now substitute the value of molecules of lead (II) iodide in formula to calculate the moles of lead (II) iodide.
Moles$= \dfrac{{4.7 \times {{10}^{22}}}}{{60.22 \times {{10}^{22}}}}$
On simplifying we get,
$\Rightarrow$ Moles = 0.0780mol
The moles of $4.7 \times {10^{22}}$ molecules is $0.0780mol$.
We know that we can convert moles to grams using the molar mass. When we multiply the molar mass with moles of the substance we would get the mass of the substance in grams.
${\text{Grams}} = {\text{Moles}} \times {\text{Molar mass}}$
We know that the molar mass of lead (II) iodide is $461g/mol$.
Let us now substitute the values of moles and molar mass of lead (II) iodide in the expression of grams.
Grams$= 0.0780\not{{mol}} \times \dfrac{{461.01g}}{{\not{{mol}}}}$
On simplifying we get,
$\Rightarrow$ Grams = 35.95g
Grams = 36.1g
The mass of $4.7 \times {10^{22}}$ molecules in grams is 36.1g.

So, the correct answer is Option A.

Note: We can convert moles to molecules using Avogadro number. Consider the example,
Example: Calculate the number of molecules $2.5mol{S_8}$.
Given,
The number of moles of ${{\text{S}}_{\text{8}}}$ is $2.5mol$
The Avogadro’s number is $6.022 \times {10^{23}}molecules$
The number of molecules can be calculated as,
$2.5mol\left( {\dfrac{{6.022 \times {{10}^{23}}molecule}}{{mol}}} \right) = 15.055 \times {10^{23}}molecule$
The number of molecules $2.5mol{S_8}$ is $15.055 \times {10^{23}}molecules$.