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

Question

Graph between log x/m and log p is a straight line inclined at an angle of 45°. 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

Tardigrade Admin · Accepted Answer

log (x/m) = log k +(1/n) log P (1/n) = tan 45° n = 1 log k = 0.693 log k = 0.3 log (x/m) = 0.3 + log (0.5) log(x/m) = 0.3 - 0.3 = 0 (x/m) = 100 = 1

Q. Graph between $l o g x / m$ and $l o g p$ is a straight line inclined at an angle of $4 5^{\circ}$ . When pressure is $0.5$ atm and $l n k = 0.693$ , the amount of solute adsorbed per gram of adsorbent will be:

$1$

$1.5$

$0.25$

$2.5$

$l o g \frac{x}{m} = l o g k + \frac{1}{n} l o g P$
$\frac{1}{n} = t an 4 5^{\circ}$
$n = 1$
$l o g k = 0.693$
$l o g k = 0.3$
$l o g \frac{x}{m} = 0.3 + l o g (0.5)$
$l o g \frac{x}{m} = 0.3 - 0.3 = 0$
$\frac{x}{m} = 1 0^{0} = 1$

Q. Graph between logx/m and logp is a straight line inclined at an angle of 45∘. When pressure is 0.5 atm and lnk=0.693, the amount of solute adsorbed per gram of adsorbent will be:

1

1.5

0.25

2.5

logmx​=logk+n1​logP n1​=tan45∘ n=1 logk=0.693 logk=0.3 logmx​=0.3+log(0.5) logmx​=0.3−0.3=0 mx​=100=1

Q. Graph between $l o g x / m$ and $l o g p$ is a straight line inclined at an angle of $4 5^{\circ}$ . When pressure is $0.5$ atm and $l n k = 0.693$ , the amount of solute adsorbed per gram of adsorbent will be:

$1$

$1.5$

$0.25$

$2.5$

$l o g \frac{x}{m} = l o g k + \frac{1}{n} l o g P$
$\frac{1}{n} = t an 4 5^{\circ}$
$n = 1$
$l o g k = 0.693$
$l o g k = 0.3$
$l o g \frac{x}{m} = 0.3 + l o g (0.5)$
$l o g \frac{x}{m} = 0.3 - 0.3 = 0$
$\frac{x}{m} = 1 0^{0} = 1$