If the initial pressure of a gas is 0.03 atm, the mass of the gas adsorbed per gram of the adsorbent is ___________ × 10⁻² g.

Solution

Chemistry ma’am · 4 steps

Read the Freundlich line: the intercept gives log k and the slope gives 1/n, then just plug in p.

1. Freundlich isotherm. Taking logs of x/m = k p^(1/n) gives a straight line. Its slope is 1/n and its intercept on the log x/m axis is log k.

   log x/m = log k + 1/n log p

2. Read the graph. The slope is 1, so 1/n = 1. The intercept is 0.602 = log 4, so k = 4.
3. Put in numbers. With 1/n = 1, the isotherm becomes x/m = k p = 4 × 0.03 = 0.12 g per gram of adsorbent.
4. So. 0.12 g = 12 × 10⁻² g, so the answer is 12.

The common slip: 0.12. Reporting x/m itself gives 0.12. The blank is the multiplier of 10⁻², so the answer is 12.

The idea to keep. In a log x/m versus log p plot, the intercept is log k, not k. Take the antilog first, then use x/m = k p^(1/n).

Answer from the NTA final answer key. Solution by PrepWiser.