1.

25cm^(3) of 0.2 M solution of metal chloride (MCl_(x)) reacted with 150cm^(3) of 0.1 M AgNO_(3) solution completely to form the precipitate of AgCl. What is the formula of metal chloride ?

Answer»

Solution :The reaction will :`""MCl_(x)+xAgNO_(3) rarr xAgCl+(NO_(3))_(x)`
`25cm^(3)" of 0.2 M "MCl_(x)" soluiton contain "25xx0.2" MILLIMOLES, i.e., 5 millimoles of "5xx10^(-3)" moles"`
`150cm^(3)" of 0.1 M "AgNO_(3)" solution CONTAINS "150xx0.1" millimoles i.e., 15 millimoles or "15xx10^(-3)" mole of "AgNO_(3)`
1 mole of `MCl_(x)` reacts with x moles of `AgNO_(3)`
`therefore 5xx10^(-3)` mole of `MCl_(x)` will react with `AgNO_(3)=5xx10^(-3)x` moles
`"As the given amounts react completely, "5xx10^(-3)x=15xx10^(-3)or x=3`
Hence, the FORMULA will be `MCl_(3)`.


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