Question

Graph between $log \,x/m$ and $log \,p$ is a straight line inclined at an angle of $45^{\circ}$. When pressure is $0.5$ atm and $ln \,k = 0.693$, the amount of solute adsorbed per gram of adsorbent will be:

• A. $1$
• B. $1.5$
• C. $0.25$
• D. $2.5$

Answer 1

Answer: (A)

$log \frac{x}{m} = log \,k +\frac{1}{n} log\,P$
$\frac{1}{n} = tan\,45^{\circ}$
$n = 1$
$log\,k = 0.693$
$log\,k = 0.3$
$log \frac{x}{m} = 0.3 + log\,(0.5)$
$log\frac{x}{m} = 0.3 - 0.3 = 0$
$\frac{x}{m} = 10^0 = 1$