Question

180
Sblem 8.34: Calculate (a) the number of atoms of carbon present in 9g
cose (b) the number of nitrate ions present in 32.8g calcium nitrate.

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Answer 1

(a) Mass of glucose = 9g

Moles of glucose = Given mass / Molar mass = 9/180 = 0.05 mol

1 mole of glucose (C6H12O6) = 6.02 x 10^23 molecules = 180g

no. of C atoms in 1 mole of glucose (C6H12O6) = 6 x 6.02 x 10^23 = 3.612 x 10^24

Atoms of C = atoms of C present in 1 mole glucose (C6H12O6) x Moles = 3.612 x 10^24 x 0.05 = 0.1806 x 10^24 = 1.8 x 10^23 atoms

(b) Mass of Ca(NO3)2 = 32.8g

Moles of Ca(NO3)2 = Given mass / Molar mass = 32.8/164 = 0.2 mol

1 mole of Ca(NO3)2 = 6.02 x 10^23 molecules = 164g

No. of NO3⁻ ions in 1 mole of Ca(NO3)2 = 2 x 6.02 x 10^23 = 1.204 x 10^24

NO3 ions present in 32.8g Ca(NO3)2 = 0.2 x 1.204 x 10^24 = 2.4 x 10^23 ions