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
A.
$$1.568 \times {10^3}$$
B.
$$1.568 \times {10^4}$$
C.
$$15.68$$
D.
$$3.136 \times {10^4}$$
Answer :
$$1.568 \times {10^4}$$
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
$$\eqalign{
& M.\,\,Wt = \frac{{100}}{{0.5}} \times 78.4 \cr
& = 1.568 \times {10^4} \cr} $$
Releted MCQ Question on Physical Chemistry >> Some Basic Concepts in Chemistry
Releted Question 1
$$27 g$$ of $$Al$$ will react completely with how many grams of oxygen?