how many grams of k2cr207are therein 8.90×1024 molecules of k2cr207
1 mole of matter contains 6.022 × 10²³ units of that substance.
n = m / M
M (K2Cr2O7) = 294.2 g/mol
n (K2Cr2O7) = 8.90×1024 / 6.022 × 10²³ = 14.8 mol
m (K2Cr2O7) = n x M = 14.8 x 294.2 = 4348 g
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