Question

180 g of glucose (C₆H₁₂O₆) contains ……………. carbon atoms.

(A) 1.8 × 10²³ 

(B) 1.8 × 10²⁴

(C) 3.6 × 10²³

(D) 3.6 × 10²⁴

Answer 1

(D) \(3.6 × 10^{24}\)

Number of moles of \(C_6H_{12}O_6=\frac{180}{180}=1\,\text{mole}\)

\(\therefore\) one molecule of \(C_6H_{12}O_6\) contain = 6 carbon atoms.

\(\therefore\) 1 mole of \(C_6H_{12}O_6\) molecule contain = \(6\times6.022\times10^{23}\) carbon atom

= \(3.6\times10^{24}\) carbon atom.

Answer 2

Option : (D) 3.6 × 10²⁴