At standard pressure the boiling point of an unsaturated nano3(aq) solution increases when
2 answers:
You might be interested in
Explanation:
The given data is as follows.
Heat of vaporization () = 45.4 kJ/mol
Specific heat () = 0.903
Let us assume the alcohol given is and its mass is 1.12 g. Also, mass of aluminium block is 73.0 g.
First, calculate the moles of alcohol () as follows.
No. of moles alcohol =
=
= 0.0186 mol
So, heat absorbed by the alcohol () = heat lost by aluminium (
)
=
12.71 =
Thus, we can conclude that the final temperature of the block is 12.3^{o}C[/tex].