Question

Anaerobically treated effluent has MPN of total coliform as ${10}^6/100mL$. After chlorination, the MPN value declines to ${10}^2/100mL$. The percent removal (%R) and log removal (log R) of total coliform MPN are

• A. $\%R=99.99;logR=2$
• B. $\%R=99.90;logR=4$
• C. $\%R=99.99;logR=4$
• D. $\%R=99.90;logR=2$

Answer 1

The correct option is C $\%R=99.99;logR=4$

Initial MPN,
$N_1={10}^6/100mL$
Final MPN,
$N_2={10}^2/100mL$

$\%\text{removal}=\frac{N_1-N_2}{N_1}\times 100$
$=\frac{{10}^6-{10}^2}{{10}^6}\times 100$
$=99.99\%$
$\text{Log removal}=log\left(\frac{N_1}{N_2}\right)=log\left(\frac{{10}^6}{{10}^2}\right)$
$=4$