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So the question is: o.o497g of a chloride of a group 1 metal called Z is dissolved on water, Excess acidified silver nitrate solution is added to the solution. The resulting precipitate is filtered and dried to constant mass. The mass of silver chloride formed is 0.957g. Deduce which metal is present?

So first I found the RMM of the silver chloride and then I found the number of moles in it using the mass given in the question. But I really don't know what to do after this part or if I even did the right thing in the first place.

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    $\begingroup$ Are your sure about your numerical values ? I doubt that such a small mass of any possible chloride can produce a mass of AgCl which is about 20 times higher ? No atoms of the periodical table can make such a chloride. $\endgroup$ – Maurice Nov 22 at 19:49
  • $\begingroup$ Welcome to Chemistry SE site. We have a policy which states that ‎you should show your thoughts, effort and attempts to understand underlying principles and solve the question. It'll make us certain that ‎we aren't doing your homework for you. ( As homework is considered literal homework, self-study questions, puzzles, worked examples etc.) Please provide your full reasoning or thoughts on this, otherwise, the question may get closed.‎ For more, see Homework $\endgroup$ – Poutnik Nov 22 at 20:49
  • $\begingroup$ Present your algebraic computation and your complete thoughts and attempts to solve it. $\endgroup$ – Poutnik Nov 22 at 20:50
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    $\begingroup$ @Maurice is right. There must be some error in the numerical vales. Even 0.0497g of $\ce{Cl-}$ anions can't make 0.957g of $\ce{AgCl}$. $\endgroup$ – MaxW Nov 22 at 21:28
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    $\begingroup$ The mistake is in the first data. If we forget the first zero in the mass of chloride, you get $0.497$ g, which leads to a correct solution. Start from $0.957$ g AgCl which contains $6.6764$ millimole AgCl. This AgCl is coming from $ 6.6764$ millimole MCl, which weighs $0.497 g$. The molar mass of MCl is$ 0.497 g/ 6.6764 mol = 76.44$. The atoms mass of M is $76.44 - 35.45 = 39.09$. This is exactly the atomic mass of potassium K ! Why was the question closed ? $\endgroup$ – Maurice Nov 23 at 12:47