Equivalent weight of a metal chloride is 75.5. How many moles of NaOH is required to completely precipitate one mole of metal hydroxide. Atomic weight of the metal is 120.

Equivalent weight of a metal chloride is 75.5. How many moles of NaOH

1.	Equivalent weight of a metal chloride is 75.5. How many moles of NaOH is required to completely precipitate one mole of metal hydroxide. Atomic weight of the metal is 120.
Answer» Solution :EQ of metal CHLORIDE = EQ wt of metal + EQ wt of `Cl^(-)` 75.5 = E + 35.5 E = 40 VALENCY = `(120)/(40)=3` `THEREFORE` Formula of metal chloride = `MCl_(3)` `MCl_(3)+3NaOHrarrM(OH)_(3)darr+3NaCl`