Question

Calculate the total number of oxygen atoms present in $1.80g$ of glucose $\left(C_6{H}_{12}{O}_6\right)$.

• A. $46.138\times {10}^{21}$
• B. $36.138\times {10}^{21}$
• C. $26.138\times {10}^{21}$
• D. $6.138\times {10}^{21}$

Answer 1

Answer: (B)

$36.138\times {10}^{21}$

$Glucose⟶C_6{H}_{12}{O}_6$

$Numberofmoles=\frac{mass}{molecularmass}$

$=\frac{1.8}{180}$

$=0.01mol$

Now, there are 6 atoms of oxygen

Total number of oxygen atoms$=$number of atoms of oxygen$\times$number of moles$\times$Avagadro's constant

$=6\times 0.01\times 6.023\times {10}^{23}$

$=36.138\times {10}^{21}$