In the balanced chemical equation for the formation of ammonia (NH3) from
1 answer:
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Explanation:
The given data is as follows.
Weight of solute = 75.8 g, Molecular weight of solute (toulene) = 92.13 g/mol, volume = 200 ml
- Therefore, molarity of toulene is calculated as follows.
Molarity =
=
= 4.11 M
Hence, molarity of toulene is 4.11 M.
- As molality is the number of moles of solute present in kg of solvent.
So, we will calculate the molality of toulene as follows.
Molality =
=
= 8.6 m
Hence, molality of given toulene solution is 8.6 m.
- Now, calculate the number of moles of toulene as follows.
No. of moles =
=
= 0.8227 mol
Now, no. of moles of benzene will be as follows.
No. of moles =
=
= 1.2239 mol
Hence, the mole fraction of toulene is as follows.
Mole fraction =
=
= 0.402
Hence, mole fraction of toulene is 0.402.
- As density of given solution is 0.857
so, we will calculate the mass of solution as follows.
Density =
0.857 =
(As 1
= 1 g)
mass = 171.4 g
Therefore, calculate the mass percent of toulene as follows.
Mass % =
=
= 44.22%
Therefore, mass percent of toulene is 44.22%.