Question

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

Answer 1

Answer:

6.68×1023 atoms.

Explanation:

Step 1). Molar mass of sucrose, C12H22O11=12×12+1×22+16×11=342g

            or 342g = 1 mole of sucrose.

         3.42 g = 0.01 mole of sucrose.

  ∴ 1 mole of sucrose (C12H22O11) contains O-atoms =11×6.022×1023 atoms

   ∴ 0.01 mole of sucrose will contain O-atoms.

    =0.01×11×6.022×1023 atoms =6.6242×1022

Step 2: 18 g of water (H2O) = 1mole of water.

            1 mole of water (H2O) contains O-atoms =6.022×1023 atoms.

Step 3: By adding the number of O-atoms present in 3.42g of sucrose and    

           18g of water, we get

           6.022×1023+6.6242×1022=1022(60.22+6.6242)  =

           66.844×1022 = 6.68×1023 atoms.