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Question: Calculate the no. of atoms of oxygen present in 88g of \(\text{C}{{\text{O}}_{2}}\). What would be t...

Calculate the no. of atoms of oxygen present in 88g of CO2\text{C}{{\text{O}}_{2}}. What would be the weight of CO\text{CO}having the same number of oxygen atoms?
A. 24.09 × 1023, 114g\text{24}\text{.09 }\times \text{ 1}{{\text{0}}^{23}}\text{, 114g}
B. 24.08 × 1025, 112g\text{24}\text{.08 }\times \text{ 1}{{\text{0}}^{25}}\text{, 112g}
C. 25.09 × 1025, 116g\text{25}\text{.09 }\times \text{ 1}{{\text{0}}^{25}}\text{, 116g}
D. 24.09 × 1023, 112g\text{24}\text{.09 }\times \text{ 1}{{\text{0}}^{23}}\text{, 112g}

Explanation

Solution

To solve the questions related to the mole concept one should know the atomic mass of each element like the atomic of carbon is 12 a.m.u and of oxygen 16 a.m.u. A number of moles is the ratio of the no. of atoms to the Avogadro's number.

Complete answer:
-It is given that the mass of carbon dioxide is 88h but we know that one of carbon dioxide has a mas of 44g i.e 12 × (16 × 2) = 44g\text{12 }\times \text{ (16 }\times \text{ 2) = 44g}.
-So, it means that there are a total of two moles of carbon dioxide.
-Now, if one mole CO2\text{C}{{\text{O}}_{2}} has 2 oxygen atoms so the number of oxygen atoms in the 2 moles of CO2\text{C}{{\text{O}}_{2}}will be: 2 × 2 × 6.022 × 1023 = 24.09 × 10232\text{ }\times \text{ 2 }\times \text{ 6}\text{.022 }\times \text{ 1}{{\text{0}}^{23}}\text{ = 24}\text{.09 }\times \text{ 1}{{\text{0}}^{23}}.
-Now, we have to calculate the number of oxygen atoms in CO, as we know that there is 1 oxygen atom in CO.
-So, the mass of CO containing 24.09 × 1023\text{24}\text{.09 }\times \text{ 1}{{\text{0}}^{23}} oxygen atoms will be:
Mass = Molar mass × No. of atoms6.022 × 1023\text{Mass = }\dfrac{\text{Molar mass }\times \text{ No}\text{. of atoms}}{\text{6}\text{.022 }\times \text{ 1}{{\text{0}}^{23}}}
Here, mass is to be calculated and no. of atoms had calculated above i.e. 24.09 × 1023\text{24}\text{.09 }\times \text{ 1}{{\text{0}}^{23}}.
-So, the mass will be: Mass = 28 × 24.09 × 10236.022 × 1023 = 112g\text{Mass = }\dfrac{28\text{ }\times \text{ 24}\text{.09 }\times \text{ 1}{{\text{0}}^{23}}}{6.022\text{ }\times \text{ 1}{{\text{0}}^{23}}}\text{ = 112g}.
-Hence, 112g of carbon monoxide will have the same number of oxygen atoms as that in carbon dioxide.
So, the correct answer is “Option D”.

Note: The Law of multiple proportions states that when two elements together combine and forms more than one compound the mass of one element which combines with a fixed of the other element, will always give the whole number ratio. That's why we can say that the mass carbon in CO and CO2\text{C}{{\text{O}}_{2}} is the same i.e. 12g.