Question

3.42 g of glucose are dissolved in 18 g of water in a beaker. The number of oxygen atoms in the solution is 

A. 6.68 × 10²³ 

B. 6.09 × 10²²  

C. 6.022 × 10²³ 

D. 6.022 × 10²¹ 

Answer 1

Molar mass of sucrose = C₁₂H₂₂O₁₁ 

= 12× 12 + 1× 22 + 16 × 11 = 342g/mol 

Number of moles = \(\frac{Mass\,of\,glucos}{Molar\,mass\,of\,sucrose}\)

⇒ Number of moles = \(\frac{3.42g}{342g/mol}\)

⇒ Number of moles = 0.01 

Sucrose (C₁₂H₂₂O₁₁) contains 11 oxygen atoms 

⇒ 11 × 6.022 × 10²³ 

For 0.01 moles of sucrose 

⇒ 0.01 × 11 × 6.022 × 10²³ = 6.6 × 10 22 

Now, Molar mass of water = H₂O = 2× 1 + 16 = 18g/mol 

Number of moles =\(\frac{Mass\,of\,water}{Molar\,mass\,of\,water}\)

⇒ Number of moles = \(\frac{18g}{18g/mol}\)

⇒ Number of moles = 1 

Sucrose (H₂O) contains 1 oxygen atom = 6.022 × 10²³ 

For 1 mole of water = 6.022 × 10²³ 

Now, add the both values: 6.6 × 10²²+ 6.022 × 10²³ We get 6.68 × 10²³ atoms. 

Hence, the option (a) is correct.