Verified Answer :
According to Avogadro's law, 1 mole ,i.e., 6.2 10^23 particles of
every substance have mass equal to its molecular mass in grams. Since
the molecular mass of CuSO4 is 80 grams. It means 6.2 10^23 particles
of CuSO4 have a mass of 80 grams. Now u can calculate the mass of 3.36
10^23 particles by unitary method 3.36 10^23 80/6.23 10^23 grams
Answer:
The weight in grams of \(3.36\times10^{23}\) molecules of copper sulfate is 89.023 g.
Step-by-step explanation:
Given : Molecules of copper sulfate \(3.36\times10^{23}\)
To find : What is the weight in grams of \(3.36\times10^{23}\) molecules of copper sulfate?
Solution :
Molar mass of copper sulfate (\(CuSO_4\))
Cu=63.5
S=32
O=16
\(CuSO_4=63.5+32+4\times16= 159.5 g\)
We know, that 1 mole = Avogadro number
1 mole = \(6.02\times 10^{23}\) molecules
\(6.02\times10^{23}\) molecule weigh = 159.5 g
1 molecule weigh = \(\frac{159.5}{6.02\times10^{23}}\) g
So, \(3.36\times10^{23}\) molecule weigh = \(\frac{159.5\times3.36\times10^23}{6.02\times10^{23}}\) g
\(=\frac{159.5\times3.36}{6.02}\) g
\(=\frac{535.92}{6.02}\) g
\(=89.023\) g
Therefore, The weight in grams of \(3.36\times10^{23}\) molecules of copper sulfate is 89.023 g.