Question

What is the weight in grams of 3.36*10^23 molecules of copper sulfate

Answer 1

                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 2

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.