Question

How many O-atoms are present in $\frac{3}{8}$ millimoles of $\left[Co\right(H_{2}O{)}_{4}C{l}_2]Cl$?

• A. $9.033\times {10}^{24}$
• B. $2.26\times {10}^{24}$
• C. $9.033\times {10}^{20}$
• D. $2.26\times {10}^{20}$

Answer 1

The correct option is C $9.033\times {10}^{20}$

1 mole of $\left[Co\right(H_{2}O{)}_{4}C{l}_2]Cl$ formula unit contains 4 moles of O-atoms.
So, $\frac{3}{8}$ millimoles of $\left[Co\right(H_{2}O{)}_{4}C{l}_2]Cl$ contain $\frac{3}{8}\times 4=\frac{3}{2}$ millimoles of O-atoms.
Again, 1 mole of O-atoms $=6.022\times {10}^{23}$ O-atoms.
Therefore, $\frac{3}{2}$ millimoles of O-atoms $=\frac{3}{2}\times {10}^{-3}\times 6.022\times {10}^{23}$ O-atoms
$=9.033\times {10}^{20}$ O-atoms