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How many grams would 3.36 × 1023 molecules of copper (II) sulfate (CuSO4) weigh?
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Divide 3.36 × 10^23 by avogadro's number (6.022 x 10^23) Then you have the amount of moles copper(II)sulfate You multiply that with the molecular weight of CuSO4 (159.6095 g/mol)
mole = number of molecules/Avogadro number mole = 3.36 x 10^23 / 6.02 x 10^23 =0.56 to find mass you multiply moles with relative molecular mass of CuSO4 mass = 0.56 x 159.5 = 89.32 grams
\[\frac{ 3.36 * 10^{23} }{ 6.022 * 10^{23} }=0.5579 mol\]0.5579 * 159.6095 = 89.0461 gram
For people on Plato looking back, Awanyulian was close but it's actually 89.03 g
@WhatEven actually... it's 89.05478579
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