What is the number of particles in 27.5 grams of chromium (iii) sulfite ?

= 4.81 × 10^22 particles

The molar mass of Cr2(SO3)3 is 344.18 g

1 mole of a substance contains 6.02 10^23 particles

But; 1 mole of chromium (iii) sulfite = 344.18 g

Therefore;

344.18 g = 6.02×10^23 particles

27.5 g = ?

Hence;

= (27.5 × 6.02 × 10^23)/344.18

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explanation: