1.

1.502 g sample of steel yields 0.259 g of nickel dimethylglyoximate NiC_(8)H_(14)N_(4)O_(4)(molar mass 289 g mol^(-1)). What is the percentage of nickel in the steel? (Atomic mass of Ni=59)

Answer»

Solution :`underset("59 g")(Ni^(2+))+underset("Dimethylglyoxime")(2C_(4)H_(8)N_(2)O_(2))rarrunderset("289 g")(NiC_(8)H_(14)N_(4)O_(4))+2H^(+)`
Thus, 289 g of glyoximate is obtained from Ni = 59 g
`THEREFORE` 0.259 g of glyoximate is obtained from Ni `= (59)/(289)xx0.259 g`
This amount of Ni is PRESENT in 1.502 g of the sample of steel.
`therefore % " of Ni in steel "= (59)/(289)XX(0.259)/(1.502)xx100=3.52`


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