NCERT Exemplar Solutions Class 9 Science Solutions for Atoms and Molecules - Exercise in Chapter 3 - Atoms and Molecules
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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
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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