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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
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Answer:
Explanation:
To calculate their average atomic masses which is otherwise known as the relative atomic mass, we simply multiply the given abundances of the atoms and the given atomic masses.
The abundace is the proportion or percentage or fraction by which each of the isotopes of an element occurs in nature.
This can be expressed below:
RAM = Σmₙαₙ
where mₙ is the mass of isotope n
αₙ is the abundance of isotope n
for this problem:
RAM of Li = m₆α₆ + m₇α₇
m₆ is mass of isotope Li-6
α₆ is the abundance of isotope Li-6
m₇ is mass of isotope Li-7
α₇ is the abundance of isotope Li-7