Question

The largest no of molecule are present in a)2.8g of CO b)3.6g of water c)4.6g of C2H5OH

Answer 1

Answer: 2. 3.6g of water.

Explanation:

No. of moles=given mass/molecular mass.

No.of molecules = no.of moles*avagadro's no i.e.6.022*10*23

So in 2.8g of CO no of moles is = 1 and no of molecules = 1*6.022*10*23

In 3.6g of water no.of moles = 2 and no of molecules=2*6.022*10*23

In 4.6g of C2H5OH no of moles is 1 so no of molecules = 1*6.022*10*23.

HOPE IT WILL HELP YOU.

Answer 2

Answer:

B 3.6g of water

Explanation:

No of molecules = given mass × NA / molar mass