Question

Calculate the number of moles of water in 488 g of bacl2, 2h20

Answer 1

molecular weight of BaCl₂.2H₂O = 244 g
given weight = 488 g

mole = given weight / molcular weight
= 488/244 = 2

Answer 2

Answer:

4 moles of water is present.

Step-by-step explanation:

First calculate the molar mass of $\mathrm{Bacl} 2.2 \mathrm{H}_{2} \mathrm{O}$ = 137 + 35 + 2(1+15) = 244 gm/mole

Each mole of the given compound contains 2 mole of $\mathrm{H}_{2} \mathrm{O}$.  

So, 244gm of $\mathrm{Bacl} 2.2 \mathrm{H}_{2} \mathrm{O}$ = 2 moles of $\mathrm{H}_{2} \mathrm{O}$

1 gm of this compound will have = 2/244 mole of $\mathrm{H}_{2} \mathrm{O}$

Therefore, 488 g of this compound will have = (2 X 488)/244 = 4 moles

488 g of the given compound having 4 moles of $\mathrm{H}_{2} \mathrm{O}$.