Question

how many grams are 5.119 x 10 ^23 molecules of copper sulfate

Answer 1

Explanation:

first divide the given no. with avogadros no. to ge no. of moles. if the no. of moles you got is n

and them M be the molar mass of CuSO4

ans= M×n

Answer 2

Given - number of molecules of copper sulfate - 5.119 x 10 ^23 molecules

Find - weight in grams of these molecules.

Answer - We are familiar with the fact that 1 mole of any compound contains number of molecules equivalent to Avogadro's number. Avogadro's number is - 6.023*10^23.

Since 6.023*10^23 molecules are present in 1 mole of copper sulphate, hence, 5.119 x 10 ^23 molecules will be present in - 0.84 moles.

Now, as we know 1 mole of copper sulphate weighs 159.6 g/mol. Then, 0.84 mole will weigh - 134.064 grams.

Hence, weigh of 5.119 x 10 ^23 molecules of copper sulphate will be - 134.064 grams.