Ask Question
28 September, 16:32

How many grams are 3.01 * 1023 molecules of CuSO4?

+1
Answers (1)
  1. 28 September, 19:04
    0
    Answer is: 79.8 grams of copper (II) sulfate.

    N (CuSO₄) = 3.01·10²³; number of molecules.

    n (CuSO₄) = N (CuSO₄) : Na.

    n (CuSO₄) = 3.01·10²³ : 6.02·10²³ 1/mol.

    n (CuSO₄) = 0.5 mol; amount of substance.

    m (CuSO₄) = n (CuSO₄) · M (CuSO₄).

    m (CuSO₄) = 0.5 mol · 159.6 g/mol.

    m (CuSO₄) = 79.8 g; mass of substance.

    M - molar mass.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “How many grams are 3.01 * 1023 molecules of CuSO4? ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers