Question

An ideal solution is prepared by mixing 15.6 g of toluene, C6H5CH3, and 136.2 g of butan-

Answer 1

This problem is providing the masses of both toluene and butan-1-ol and the vapor pressure of the latter when pure. Thus, it asks for the vapor pressure of this alcohol in the mixture, therefore the answer turns out 812 Pa.

Raoult's law

In chemistry, when two pure liquids are mixed, their properties as a mixture differ from those when pure, for that reason we use the Raoult's law in order to quantify these properties as pure substances, or when mixed with other substances.

In such a way, one can write this law for butan-1-ol as follows:

$P_{but}=P_{but}^{pure}*x_{but}$

Where we know its vapor pressure as 885 Pa and we want to calculate its partial pressure but first its molar fraction with the given masses converted to moles as follows:

$x_{but}=\frac{n_{but}}{n_{but}+n_{tol}} \\\\n_{but}=136.2g*\frac{1mol}{74.121g}=1.838mol\\ \\n_{tol}=15.6g*\frac{1mol}{94.14g} =0.166mol\\\\x_{but}=\frac{1.838mol}{1.838mol+0.166mol}=0.927$

Finally, we solve for its partial pressure:

$P_{but}=885Pa*0.917\\\\P_{but}=812Pa$

Learn more about the Raoult's law: brainly.com/question/12969696