Graph between log $\frac{x}{m}$ and log $P$ is a straight line inclined at an angle $45^{o}$ . When pressure of $0.5 a t m$ and $l o g k = 0.699,$ the amount of solute adsorbed per $g$ of adsorbent will be (Given, antilog 0.699 = 5)

1 g/g adsorbent

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1.5 g/g adsorbent

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2.5 g/g adsorbent

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0.25 g/g adsorbent

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Solution

The correct option is C $2.5 g/g adsorbent$
$\frac{x}{m} = k . P^{1 / n} since log k = 0.699$
$Hence k = 5$
$Slope = \frac{1}{n} = t a n 45^{o} = 1, thus, \frac{x}{m} = 5 \times 0.5 = 2.5 g/g adsorbent$

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Similar questions

Q. Graph between log

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θ = 45^{\circ}

. When pressure is 0.5 atm and log k = 0.699, the amount of solute adsorbed per g of adsorbent will be:

Graph between $log \frac{x}{m}$ and $log P$ is a straight line inclined at an angle $θ = 45^{\circ}$ . When pressure is 0.5 atm and log K = 0.699, the amount of solute adsorbed per g of adsorbent will be:

Plot of log x/m against log P is a straight line inclined at an angle of $45^{\circ}$ . When the pressure is 0.5 atm and Freundlich parameter. K is 100, the amount of the solute adsorbed per gram of adsorbent will be (log 5 = 0.6990)

Q. The plot of

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against

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is a straight line inclined at an angle of

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Graph between log xm and log P is a straight line inclined at an angle 45o. When pressure of 0.5 atm and log k=0.699, the amount of solute adsorbed per g of adsorbent will be (Given, antilog 0.699 = 5)

The correct option is C 2.5 g/g adsorbentxm=k.P1/n since log k =0.699 Hence k=5 Slope=1n= tan45o=1, thus, xm=5×0.5=2.5 g/g adsorbent

Graph between log $\frac{x}{m}$ and log $P$ is a straight line inclined at an angle $45^{o}$ . When pressure of $0.5 a t m$ and $l o g k = 0.699,$ the amount of solute adsorbed per $g$ of adsorbent will be (Given, antilog 0.699 = 5)

The correct option is C $2.5 g/g adsorbent$
$\frac{x}{m} = k . P^{1 / n} since log k = 0.699$
$Hence k = 5$
$Slope = \frac{1}{n} = t a n 45^{o} = 1, thus, \frac{x}{m} = 5 \times 0.5 = 2.5 g/g adsorbent$