Question

How many grams are present in -
(i) 3*10^12 mole of calcium carbonate.
(ii) 2.02*10^11 mole of iron sulphate.

Answer 1

Explanation:

$\huge\underline{\underline{\sf ANSWER}} \\ for \: calcium \: carbonate = > \\ mole = \frac{given \: weight}{molecular \: weight} \\ 3 \times {10}^{12} = \frac{x}{100} \\ x = 3 \times {10}^{14} \: grams \\ here \: x \: is\: weight \: of \: Calcium \: Carbonate\\ \\ for \: iron \: sulphate = > \\ mole = \frac{given \: weight}{molecular \: weight} \\ 2.02 \times {10}^{11} = \frac{x}{152} \\ 3.0704 \times {10}^{13} \: grams = x \\ here \: x \: is \: weight \: of \: Iron \: Sulphate$