JEE Class main Answered

sir ye 4th wala kaise hua?

Asked by gnimit146 | 13 Feb, 2021, 00:53: AM

Expert Answer

Al = 26.98 g mol^–1

Let the mass of Al in the sample be x g.

Moles of straight H subscript 2 space formed equals space fraction numerator 415 space cm 3 space over denominator 22 space 414 space cm 3 space divided by space mol end fraction space equals 0.018515 mol space space space space Al space space space space space plus space space space space 3 straight H to the power of plus space space space rightwards arrow with blank on top space space space space space Al blank to the power of 3 plus end exponent space space space space space plus space space bevelled 3 over 2 space space straight H subscript 2 26.98 space straight g space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 1.5 space mol space space straight x space straight g space space space space space space space space space space space space space space space space space space space space space space space space space space space space 0.018515 space mol Taking simple ratios colon fraction numerator 26.98 over denominator straight x end fraction space equals space fraction numerator 1.5 over denominator 0.018515 end fraction straight x space equals space fraction numerator 26.98 space cross times space 0.018515 over denominator 1.5 end fraction space equals space 0.333 space straight g space Al Therefore comma the mass percent sign of Al in the sample of alloy is space fraction numerator 0.333 over denominator 0.350 end fraction space cross times 100 percent sign space equals 95.2 percent sign

Answered by Ramandeep | 15 Feb, 2021, 15:59: PM

Please Login or Sign up to continue!

JEE Class main Answered

Choose Filters

Open In Ask A Doubt App ?

Get in Touch