At temperature 63.50 °c, the vapour pressures of pure h20 and pure ethanol, c2h5oh
are 175.00 and 4.00 x 102 torr, respectively. a solution is prepared by mixing equal masses
of h20 and c2h5oh.
a) calculate the mole fraction of ethanol in the solution.
b) what is the vapour pressure of the solution at 63.50 °c?
c) determine the mole fraction of ethanol in the vapour phase above the solution.
( 0.28, 238.20 torr, 0.47)

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

answersilicon (si); oxygen (o);