The percentage of Se in peroxidase anhydrous enzyme is $0.5 \%$ by weight (atomic weight $=78.4$ ). Then…

The percentage of Se in peroxidase anhydrous enzyme is $0.5 \%$ by weight (atomic weight $=78.4$ ). Then minimum molecular weight of peroxidase anhydrous enzyme is
  1. $1.568 \times 10^{3}$
  2. $1.568 \times 10^{4}$
  3. $15.68$
  4. $3.136 \times 10^{4}$

Solution

$0.5 \%$ by weight means if Mol. wt. is 100 then mass of Se is $0.5$. If at least one atom of Se is present in the molecule then
M. $\mathrm{Wt}=\frac{100}{0.5} \times 78.4=1.568 \times 10^{4}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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