Question

Calculate the number of atoms of oxygen present in 88 g of ${CO}_2$. What would be the weight of $CO$ having the same number of oxygen atoms?

• A. $24.09\times {10}^{23},114g$
• B. $24.08\times {10}^{25},112g$
• C. $25.09\times {10}^{23},116g$
• D. $24.09\times {10}^{23},112g$

Answer 1

Answer: (D)

$24.09\times {10}^{23},112g$

As we know, molecular weight of $C{O}_2=44g$

So, number of moles of $C{O}_2=\frac{88}{44}=2$ moles and each mole of $C{O}_2$ contain 2 moles of oxygen atoms.

So, number of oxygen atoms $=4$ mol $=24.09\times {10}^{23}$ atoms.

(As one mole contain $6.022\times {10}^{23}$ atoms )

Also, only one atom of oxygen is present in each mole of $CO$ molecule.

So, weight of 4 moles of $CO=4\times 28=112g$.
(As both molecule have same no. of oxygen atoms)