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Byju's Answer

Standard XII

Physics

Vector Addition

Graph between...

Question

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:

0.397 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
$l o g \frac{x}{m} = l o g k + \frac{1}{n} l o g P l o g \frac{x}{m} = 0.699 + (t a n 45^{\circ}) (l o g (0.5)) l o g \frac{x}{m} = 0.3979 = 0.40 \frac{x}{m} = 2.5$
x=2.5 g/g adsorbent

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Similar questions

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:

Q. 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)

Q. When a graph is plotted between

log \frac{x}{m}

and

log p

, it is a straight line with an angle

45^{\circ}

and intercept

0.3010

y -

axis. If initial pressure is

0.3 atm

. what will be the amount of gas adsorbed per

g

of adsorbent:-

Q. Graph between

log (\frac{x}{m})

and

log P

is a straight line at an angle

45^{\circ}

with intercept on

y

-axis

0.3010

. The amount (in

g

) of the gas absorbed per

g

of the adsorbent when pressure is

0.2 a t m

is:
(Assume that the adsorption obey Freundlich isotherm)

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)

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Graph between log Xm and log P is a straight line inclined at an angle θ=45∘. When pressure is 0.5 atm and log k = 0.699, the amount of solute adsorbed per g of adsorbent will be:

The correct option is C 2.5 g/g adsorbentlogxm=logk+1nlogPlogxm=0.699+(tan45∘)(log(0.5))logxm=0.3979=0.40xm=2.5 x=2.5 g/g adsorbent

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:

The correct option is C 2.5 g/g adsorbent
$l o g \frac{x}{m} = l o g k + \frac{1}{n} l o g P l o g \frac{x}{m} = 0.699 + (t a n 45^{\circ}) (l o g (0.5)) l o g \frac{x}{m} = 0.3979 = 0.40 \frac{x}{m} = 2.5$
x=2.5 g/g adsorbent