NTA Abhyas 2020
- Exam
- NTA Abhyas
- Year
- 2020
- Subject
- Chemistry
- Chapter
- Chemical Equilibrium
NTA Abhyas 2020 · Chemistry · Chemical Equilibrium
Plot of (x/m) against log P is a straight line inclined at an angle...
Plot of \log (x/m) against log P is a straight line inclined at an angle When the pressure is 0.5 atm and Freundich parameter k is 10, the amount of solute adsorbed per gram of adsorbent will be (\log 5 = 0.699)
Options
- A1 g
- B2 g
- C3 g
- D5 g
Correct answer
D. 5 g
Step-by-step solution
\text\log\left(\text{x/m}\right)\text{=}\text\log\text{k}\text{+}\text{1/n}\text\log\text{P}
\text{1/n}\text{=}\text\tan{\text{45}^{o} \text{=}\text{1;}\text\log\text{k}\text{=}\text\log\text{10}\text{=}\text{1}
\text{x/m}\text{=}{\text{kp}^{\dfrac{1}{n}} \text{=}\text{10*}\left(\text{0}\text{.5}\right)\text{1}\text{=}\text{5}
When
\text{1/n}\text{=}\text\tan{\text{45}^{o} \text{=}\text{1;}\text\log\text{k}\text{=}\text\log\text{10}\text{=}\text{1}
\text{x/m}\text{=}{\text{kp}^{\dfrac{1}{n}} \text{=}\text{10*}\left(\text{0}\text{.5}\right)\text{1}\text{=}\text{5}
When