A mixturecontains NaCland an unknownchloride MCI. (i) 1 g of thisis dissolved in water . Excess ofacidified AgNO_(3) solution is added to it 2.567 g ofwhite precipitateis formed (ii) 1- 0 gof theoriginal mixtureis heated of 300^(@) C. Some vapours comeout whichare absorbed in acidified AgNO_(3) solution . 1.314 g of a whiteprecipitate is obtained.Findthe molecular weightof the unknownchloride.

A mixturecontains NaCland an unknownchloride MCI. (i) 1 g of thisis di

1.	A mixturecontains NaCland an unknownchloride MCI. (i) 1 g of thisis dissolved in water . Excess ofacidified AgNO_(3) solution is added to it 2.567 g ofwhite precipitateis formed (ii) 1- 0 gof theoriginal mixtureis heated of 300^(@) C. Some vapours comeout whichare absorbed in acidified AgNO_(3) solution . 1.314 g of a whiteprecipitate is obtained.Findthe molecular weightof the unknownchloride.
Answer» Solution :Supposethe molecular weight of MCI is M . Given that : `{:(NaCl = MCI overset(AgNO_(3))to AgCl),("(1 - x) g x g (say)2 . 567 g "):}` Applying POAC for CL atoms, moles of Clin NaCl + moles of Cl in MCI = moles of Cl in AgCl ` 1 xx ` moles of NaCl `+ 1 xx ` molesof MCl= `1 xx ` molesof AgCl `(1 - x)/( 58 . 5) + (x)/( M) = (2 . 567)/( 143 . 5 )`. . . (i) Now , futher of `300^(@) C ` MCIis supposed to undergo sublimation whileNaCldoesnot. We have, `{:(MCI underset(300^(@)C)overset(AgNO_(3))to AgCl),("x g1.341 g "):}` ApplyingPOAC for Cl atoms, moles of Cl in MCI= moles of Cl in AgCl `1 xx ` moles of MCI `= 1 xx ` molesof AgCl `(x)/( M) = (1 . 341)/( 143 . 5) "". . . (ii)` From equation (i) and (ii) , we get M = 53.5 `:.`mol. wt. of MCI= 53.5