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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 + MNO3mass of 1 mole of metal is given by dulong petite lawat wt*sp heat=25at wt=25/.057=438.6g/molenow mole of AgNO3=.51/169.87=.003mole of MClx=.22/438.6+35.5xboth are equal solve and get the answer
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