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Atoms and Molecules | Atoms and Molecules - Exercise

Question 9

3.42 g of sucrose is dissolved in 18g 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

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The answer is (a) 6.68 × 1023

Explanation:

1 mol of sucrose ( C12H22O11) contains = 11× NA atoms of oxygen, where NA = 6.023×1023

0.01 mol of sucrose (C12 H22 O11) contains = 0.01 × 11 × NA atoms of oxygen

= 0.11× NA atoms of oxygen

= 18 g/(1×2+ 16)gmol-1

=18 g /18 gmol-1

= 1mol

1mol of water (H2O) contains 1×NA atom of oxygen

Total number of oxygen atoms =

Number of oxygen atoms from sucrose + Number of oxygen atoms from water

= 0.11 NA + 1.0 NA = 1.11NA

Number of oxygen atoms in solution = 1.11 × Avogadro’s number

= 1.11 × 6.022 ×10”23 = 6.68 × 1023

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subject-cta

Question 9

3.42 g of sucrose is dissolved in 18g 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

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

The answer is (a) 6.68 × 1023

Explanation:

1 mol of sucrose ( C12H22O11) contains = 11× NA atoms of oxygen, where NA = 6.023×1023

0.01 mol of sucrose (C12 H22 O11) contains = 0.01 × 11 × NA atoms of oxygen

= 0.11× NA atoms of oxygen

= 18 g/(1×2+ 16)gmol-1

=18 g /18 gmol-1

= 1mol

1mol of water (H2O) contains 1×NA atom of oxygen

Total number of oxygen atoms =

Number of oxygen atoms from sucrose + Number of oxygen atoms from water

= 0.11 NA + 1.0 NA = 1.11NA

Number of oxygen atoms in solution = 1.11 × Avogadro’s number

= 1.11 × 6.022 ×10”23 = 6.68 × 1023

Our top 5% students will be awarded a special scholarship to Lido.

subject-cta
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