A sample of metal chloride weighing 0.22 g required 0.51 g of AgNo3 to precipitate the chloride completely. Tha specific heat of the metal is 0.057. Find the molecular formulaof the chloride if the metal is M.

MClx + AgNO3 -----AgCl + MNO3
mass of 1 mole of metal is given by dulong petite law
at wt*sp heat=25
at wt=25/.057=438.6g/mole

now mole of AgNO3=.51/169.87=.003
mole of MClx=.22/438.6+35.5x
both are equal solve and get the answer