1.

A currentof 6 amperesis passedthroughA1C1_(3) solutionfor 15 minutesusingPtelectrodes , when0.50 g A1 isproduced.Whatis the molarmass of A1 ?

Answer»


Solution :Given : Electriccurrent= I =6 A
TIME= t=15 min= 155 `xx 60 s = 900 s`
Massof A1 produced = 0.504 g
Molarmassof A1 = ?
Reductionhalfreaction
`A1_((ag))^(3)+ 3E^(-) to A1_((aq))`
Quantityof electricitypassed= Q = `I xx t`
`=6xx 900 = 5400 C`
Number a of molesof electrons `= (Q)/(F ) =(5400)/( 96500) = 0.05596mol`
Fromhalfreaction
:'3 moles of electronsdeposit1 MOLE A1
`:. 0.05596 ` molesof electronswilldeposit
`(0.05596)/(3) =0.01865mol A1 `
Now
`:' 0.01865 ` mol A1weighs0.504 g
`:.1 ` moleA1 willweigh ,` (0.504)/(0.01865)= 27 g`
Hencemolarmass ofA1 is27 g `mol^(-1)`


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