[tex]6.023\text{ x 10}^{23}\text{ molecules are present in 1 mole of CuSO}_4\text{ i.e. 159.5 g of CuSO}_4[/tex]
So, using unitary method
[tex]3.36\text{ x 10}^{23}\text{ molecules will be there in }\frac{159.5}{6.023\text{ x 10}^{23}}\text{ x 3.36 x10}_{\placeholder{⬚}^{\placeholder{⬚}}}^{23}[/tex][tex]=\text{ 88.97 g of CuSO}_4[/tex]
