The passage of electric current through silver nitrate and a metal sulphate deposited 4.04 g of Silver and 1.19 g of metal. If the specific heat of metal is 0.095, calculate the atomics mass of metal.

Asked by suryansudash5 | 20th Nov, 2015, 06:41: PM

Expert Answer:

begin mathsize 11px style straight w subscript 1 over straight w subscript 2 space equals space straight E subscript 1 over straight E subscript 2 fraction numerator 4.04 over denominator 1.19 end fraction space equals space fraction numerator 108 over denominator Eq. space wt end fraction bold Eq bold. bold wt space equals space fraction numerator 108 space cross times space 1.19 over denominator 4.04 end fraction space equals space 31.81 Atomic space mass space cross times space Specific space heat space equals space 6.4 space left parenthesis approx right parenthesis bold Atomic bold space bold mass space equals space fraction numerator 6.4 over denominator 0.095 end fraction space equals space 67.37 space left parenthesis approx right parenthesis bold Valency space equals space fraction numerator Approx space atomic space wt over denominator Eq. wt end fraction space equals space fraction numerator 67.37 over denominator 31.81 end fraction space equals space 2.11 space almost equal to space 2 bold Exact bold space bold wt bold space equals space Eq. space wt space cross times Valency space equals space 31.81 space cross times space 2 space equals space 63.62 end style

Answered by Arvind Diwale | 23rd Nov, 2015, 05:16: PM

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