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See it's an easy one..!!

AgNO3 + HCl -> AgCl + HNO3

100 mL of 0.068 M AgNO3 contains AgNO3 = 0.068 mol

So,

25 mL of 0.068 M AgNO3 contains AgNO3 = (0.068 * 25) / 1000 = 0.0017 mol

From the equation, we can see

1 mol of AgNO3 gives 1 mol of AgCl

0.0017 mol of AgNO3 gives 0.0017 mol of AgCl

Amount of AgCl can be found this way.!

No. of moles = given mass/ molecular mass

molecular mass of AgCl = 107+35.5 = 143.5 g

Therefore,

Given mass = No. of moles*molecular mass = 0.0017*143.5 = 0.244g

Note : In your question, you have written 0.068 "m" .. (small) m represents for Molality and (capital) M represents for Molarity..!

Hope I helped.. :)

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Q: What mass of solid agcl is obtained when 25 ml of 0.068 m agno3 reacts with excess of aqueous hcl?
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