3.42 gm of sucrose are dissolved in a beaker the number of oxygen atom is

Open in App

Solution

You didn't specified the amount of water present in your question. Oxygen is present in water also. So I am just taking amount of water is 18 gram for calculation.

C12H22O11 + H2O → 2C6H12O6

3.42 gm 18gm

0.1mole 1mole

0.1 mole of sucrose contain = 0.1 x 6.023 x 10^23sucrose molecules = 11 x 0.6023 x 10^23 oxygen atoms

1 mole of water contain = 6.023 x 10^23 water molecules = 1 x 6.023 x 10^23 oxygen atoms

Total number of oxygen atoms = 11 x 0.6023 x 10^23 + 1 x 6.023 x 10^23 = 12.6483 x 10^23 atoms.

Suggest Corrections

Similar questions

Q. 3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution are:

Q. 3.42 g of sucrose are dissolved in 18 g of water in a beaker. The number of oxygen atoms in the solution are
(a)

6.68 \times 10^{23}

(b)

6.09 \times 10^{22}

(c)

6.022 \times 10^{23}

(d)

6.022 \times 10^{21}

Q. 3.42g of sucrose is dissolved in 18g

H_{2} O

in a beaker. The number of oxygen atoms in this solution is:

Question 9
3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution are
(a) $6.68 \times 10^{23}$
(b) $6.09 \times 10^{22}$
(c) $6.022 \times 10^{23}$
(d) $6.022 \times 10^{21}$

Join BYJU'S Learning Program