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The graph of  \log \frac{x}{m}$ vs $\log \mathrm{p} for an adsorption process is a straight line inclined at an angle of 45^{\circ} with intercept equal to 0.6020 . The mass of gas adsorbed per unit mass of adsorbent at the pressure of 0.4 \mathrm{~atm} is  _________ \times 10^{-1} (Nearest integer)
Given: \log 2=0.3010

Option: 1

16


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \text { Slope }=\tan 45^{\circ}=1 \\ & \log \mathrm{K}=0.6020=\log 4 \\ & \mathrm{~K}=4 \\ & \frac{\mathrm{x}}{\mathrm{m}}=\mathrm{KP}^{1 / \mathrm{n}} \\ & \frac{\mathrm{x}}{\mathrm{m}}=4(0.4) 1=16 \times 10^{-1} \end{aligned}

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