Some Basic Concepts of Chemistry - Result Question 36
####35. Percentage of Se in peroxidase anhydrase enzyme is $0.5 %$ by weight (at. wt. $=78.4$ ) then minimum molecular weight of peroxidase anhydrase enzyme is
(a) $1.568 \times 10^{3}$
(b) 15.68
(c) $2.136 \times 10^{4}$
(d) $1.568 \times 10^{4}$
[2001]
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Solution:
- (d) Suppose the mol. wt. of enzyme $=x$
Given $100 g$ of enzyme wt of $Se=0.5 g$
$\therefore$ In $x$ g of enzyme wt. of $Se=\frac{0.5}{100} \times x$
Hence $78.4=\frac{0.5 \times x}{100}$
$\therefore x=15680=1.568 \times 10^{4}$ (c) Table for empirical formula :
\begin{tabular}{lllll} \hline Element $%$ & \begin{tabular}{l} At. \\ wt \end{tabular} & \begin{tabular}{l} Relative \\ number \end{tabular} & Ratio \\ \hline $C$ & 40 & 12 & $\frac{40}{12}=3.33$ & $\frac{3.33}{3.33}=1$ \\ \hline $H$ & 6.66 & 1 & $\frac{6.66}{1}=6.66$ & $\frac{6.66}{3.33}=2$ \\ \hline $O$ & 53.34 & 16 & $\frac{53.34}{16}=3.33$ & $\frac{3.33}{3.33}=1$ \\ \hline \end{tabular}
(% of $O$ in organic compound
$=100-(40+6.66)=53.34 %)$
$\therefore$ Empirical formula of organic compound $=CH_2 O$.
