CH$$_4$$ is adsorbed on 1 g charcoal at 0°C following the Freundlich adsorption isotherm. 10.0 mL of CH$$_4$$ is adsorbed at 100 mm of Hg, whereas 15.0 mL is adsorbed at 200 mm of Hg. The volume of CH$$_4$$ adsorbed at 300 mm of Hg is $$10^x$$ mL. The value of x is _________ $$\times 10^{-2}$$. (Nearest integer)
[Use $$\log_{10} 2 = 0.3010$$, $$\log_{10} 3 = 0.4771$$]

For adsorption on a solid surface the Freundlich isotherm is used, which in its usual form is written as

$$\frac{x}{m}=kP^{\,1/n}$$

Here $$x/m$$ is the amount adsorbed per gram of adsorbent. In this question the experimentally measured volume $$V$$ of CH$$_4$$ (in mL) plays the role of $$x/m$$, so we may write

$$V=k\,P^{\,1/n}$$

Taking common (base-10) logarithms on both sides, we obtain the linear equation

$$\log V=\log k+\frac{1}{n}\,\log P$$

Two data points are given:

At $$P_1=100\,$$mm Hg, $$V_1=10\,$$mL.
At $$P_2=200\,$$mm Hg, $$V_2=15\,$$mL.

We first evaluate the logarithms needed.

For the first point:

$$\log V_1=\log 10=1$$ $$\log P_1=\log 100=2$$

For the second point:

$$\log V_2=\log 15=\log 3+\log 5 =0.4771+0.6990 =1.1761$$

$$\log P_2=\log 200=\log(2\times100) =\log 2+\log 100 =0.3010+2 =2.3010$$

Substituting each set of values into $$\log V=\log k+\dfrac{1}{n}\log P$$ gives two simultaneous equations:

$$1=\log k+\frac{1}{n}(2)\quad\text{(A)}$$

$$1.1761=\log k+\frac{1}{n}(2.3010)\quad\text{(B)}$$

Now we subtract (A) from (B) to eliminate $$\log k$$:

$$1.1761-1=\frac{1}{n}(2.3010-2)$$ $$0.1761=\frac{1}{n}(0.3010)$$

Hence

$$\frac{1}{n}=\frac{0.1761}{0.3010}=0.585\;(\text{to }3\text{ dp})$$

Substituting this value into equation (A) to find $$\log k$$:

$$1=\log k+2\left(0.585\right)$$ $$1=\log k+1.17$$ $$\log k=1-1.17=-0.17$$

Now we calculate the volume adsorbed at $$P_3=300\,$$mm Hg. First,

$$\log P_3=\log 300=\log(3\times100)=\log 3+\log 100 =0.4771+2 =2.4771$$

Using $$\log V_3=\log k+\frac{1}{n}\log P_3$$ we get

$$\log V_3=-0.17+0.585\,(2.4771)$$

Multiplying:

$$0.585\times2.4771=1.4491035$$

Thus

$$\log V_3=-0.17+1.4491035=1.2791035\approx1.2791$$

Therefore

$$V_3=10^{1.2791}\ \text{mL}$$

The problem statement writes this as $$V_3=10^{x}\,$$mL, so $$x\approx1.2791$$. It then asks for the nearest integer to $$x\times10^{-2}$$, i.e.

$$x\times10^{2}=1.2791\times100=127.91$$

The nearest integer to $$127.91$$ is $$128$$.

So, the answer is $$128$$.