Answer:
The volume (in ml) of a 3.57 M solution of [tex]ZnBr_2[/tex] that contains 0.884 moles of [tex]ZnBr_2[/tex] is 247.61g mL
Explanation:
According to the definition of Molarity
which sates
"The amount of moles of a substance present in one litre of the solution is called as Molarity."
3.57M [tex]ZnBr_2[/tex] solution implies that 1000 mL of the solution conatins 3.57Moles of [tex]ZnBr_2[/tex]
Hence , 3.57 moles of [tex]ZnBr_2[/tex] is present in the 1000 ml of solution
Now,
In 1 mole of [tex]ZnBr_2[/tex] is present in the =[tex]\frac{1000}{3.57}[/tex]ml of solution
Similarily ,
In 0.884 moles of[tex]ZnBr_2[/tex] is present in the [tex]\frac{1000\times 0.884}{3.57}[/tex] of solution= 247.61g mL