Question

A person drinks $200$ g of $3$% glucose solution (molecular wt. $=180$) after aerobic exercise. The number of carbon atoms consumed by him is :

• A. $3.01\times {10}^{22}$
• B. $1.0\times {10}^{22}$
• C. $2.01\times {10}^{22}$
• D. $1.20\times {10}^{23}$

Answer 1

The correct option is D $1.20\times {10}^{23}$

Weight $=200\times \frac{3}{100}=6g$
Moles (n)$=\frac{wt}{m_w}=\frac{6}{180}=\frac{1}{30}=\frac{N}{N_A}$
Number of molecules $\left(N\right)=\frac{N_A}{30}$
$1$ molecule of glucose contain $=6C$ atom
$\frac{N_A}{30}$ molecules of glucose contain$=6\times \frac{6.023\times {10}^{23}}{30}$$=1.2\times {10}^{23}C$ atoms