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Sabina Setty
Asked: 2022-11-02T11:58:53+05:30 In:

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 `1n 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`

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 `1n 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`
Parabola
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  1. Correct Answer – A
    `”log” (x)/(M) = log K + (1)/(n) log P`
    `(1)/(n) = tan 45^(@) ” ” “In ” k = 0.69`
    `n = 1 ” ” k = 2`
    `(x)/(m) = 2 xx (0.5)^(1)`
    x = 1

    • 0
Nilam Thakkar
Asked: 2022-10-30T20:25:25+05:30 In:

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 `1n 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

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 `1n 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
Parabola
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Leave an answer

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1 Answer

  1. Correct Answer – A
    Log`x/M = log k + 1/n log P`
    `1/n = tan 45^2` 1n `k = 0.69`
    `n=1 k=2`
    `x/m = 2xx(0.5)^(1)`
    `x = 1`.

    • 0
Fakaruddin Pranay Agate
Asked: 2022-10-29T01:12:50+05:30 In:

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 `1n 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`

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 `1n 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`
Parabola
Leave an answer

Leave an answer

Browse

1 Answer

  1. Correct Answer – A
    logx/M=log K+1/n log P`
    `1/n=tan 45^(@)” “ln k=0.69`
    `n=1″ “k=2`
    `x/m=2xx(0.5)^(1)`
    `x=1`

    • 0