Question

find the number of atoms of oxygen present in 3.42 g of same sugar (C6H12O11)​

Answer 1

number of molecules =
(molecular wt. of sugar÷3.42) × 6.022 ×10^23

Answer 2

Answer:

$\sf 8.67 \times 10^{22} or\ 0.144 \times N_A$ (approx)

Explanation:

Given weight of the sugar $\sf C_6H_{12}O_{11}$ = 3.42 gram

Molar mass of $\sf C_6H_{12}O_{11}$ = 12(6) + 12 + 16(11) = 260 gram

So, mole = $\sf \frac{given \ mass}{molar \ mass}$

$\sf\implies \frac{3.42}{260} = \frac{1.71}{130}$

Now, number of atoms of oxygen in given sugar would be 11 × Avogadro Number × number of moles.

(since, number of atoms of oxygen in given sugar is 11)

So, number of atom in 3.42 gram of $\sf C_6H_{12}O_{11}$ would be

$\sf \frac{1.71}{130} \times 11 \times 6.022 \times 10^{23}$

= $\sf 8.67 \times 10^{22} or\ 0.144 \times N_A (approx)$