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Given 645 of compound C6H5N3O6 determine no. of N atoms present in compound. Report x, if your answer is x \(\times\) 1021 (C = 12, H = 1, N = 1, O = 16) (NA = 6.02 × 1023)

Answer»

Moles of compound (C6H5N3O6) = \(\frac{645}{215}\) = 3 mol 

moles of Nitrogen = 9 mole

No. of atoms of Nitrogen = 9 x 6.02 × 1023

= 54.18 × 1023

= 5418 × 1021



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