1.

A mixture of aluminium and zincweighning 1.67g was completely dissolvedin acidand evolved 1.69 litres of hydrogen at NTP . What was the aluminium in the original mixture ? (Al = 27, Zn = 65.4)

Answer»

Solution :Since `H_(2)` is formed by both Al and Zn ,
EQ. of Al + eq of Zn = eq. of `H_(2)` .
Let w be the mass in grams of Al in the MIXTURE.
`:. (w)/("eq. wt . of Al") + ((1.67 - w))/(" eq . wt .of Zn") = (1.69)/( " vol. of 1 eq. of " H_(2)" at NT in lit")`
` OMEGA /(27//3) +(1.67 - omega)/(65.4//2) = (1.69)/(11.2){:{("eq . wt of Al"=27/3),("eq. wt. of Zn"=(65.4)/2 ):}}`
` :."" omega = 1.24 g `


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