Answer:
Vapour pressure of benzene over the solution is 253 torr
Explanation:
According to Raoult's law for a mixture of two liquid component A and B-
vapour pressure of a component (A) in solution = 
vapour pressure of a component (B) in solution = 
Where
are mole fraction of component A and B in solution respectively
are vapour pressure of pure A and pure B respectively
Here mole fraction of benzene in solution is 0.340 and vapour pressure of pure benzene is 745 torr
So, vapour pressure of benzene in solution = 
= 253 torr
<h3><u>Answer;</u></h3>
Empirical formula = C₂H₃O
Molecular formula = C₁₄H₂₁O₇
<h3><u>Explanation</u>;</h3>
Empirical formula
Moles of;
Carbon = 55.8 /12 = 4.65 moles
Hydrogen = 7.04/ 1 = 7.04 moles
Oxygen = 37.16/ 16 = 2.3225 moles
We then get the mole ratio;
4.65/2.3225 = 2.0
7.04/2.3225 = 3.0
2.3225/2.3225 = 1.0
Therefore;
The empirical formula = <u>C₂H₃O</u>
Molecular formula;
(C2H3O)n = 301.35 g
(12 ×2 + 3× 1 + 16×1)n = 301.35
43n = 301.35
n = 7
Therefore;
Molecular formula = (C2H3O)7
<u> = C₁₄H₂₁O₇</u>
326000 mm is incorrect as it is either the one in cm or mm. cm aligns with m so it must be mm