Answer:
The answers to the question are as follows
In the solution
The molarity of toluene is 0.4114 M
The molality of toluene is 0.48 molal
The mass percent of toluene is 4.4224 %
The mole fraction of toluene is 0.402.
Step-by-step explanation:
The molar mass of toluene = 92.13 g/mol
The molar mass of benzene = 78.11 g/mol
The mass of toluene in the sample = 75.8 g
The mass of benzene in the sample = 95.6 g
(i) The number of moles, n of each, toluene an benzene is given by
n = Mass/(Molar Mass) = n_Toluene = 75.8 g/(92.13 g/mol) = 0.823 moles
n_benzene = 95.6 g/(78.11 g/mol) = 1.224 moles
Volume of solution = 2.00 x 10³ mL = 2 liters
Therefore 2 liters of solution contains 0.823 moles of toluene
1 liter of solution will contain (0.823 moles)/(2.L) = 0.4114 moles
and the molarity of toluene in the solution is 0.4114 M.
(ii) The molality is the number of moles per kilogram of solvent given by
molality = (Number of moles of solute)/(1 kg of solvent)
Mass of solvent = Density × Volume
Where;
Density = Density of the solution = 0.857 g/cm³
Volume = Volume of the solution = 2.00 x 10³ mL = 2 liters = 2000 cm³
Therefore mass of solvent = 0.857 g/cm³ × 2000 cm³ = 1714 g
1 kg = 1000 g ∴ 1714 g = 1.714 kg
Number of moles of toluene in 1.714 kg of solvent = 0.823 moles
We divide both sides by 1.714 kg to arrive at
The number of moles of toluene in 1.0 kg of solvent = (0.823 moles)/1.714 = 0.48 molal.
(iii) The mass percent is given by
(Mass of component/ (total mass of the mixture) ) × 100 %
The total mass of the solution ±was found as 1714 g
The mass of toluene is given as 75.8 g
The mass percent therefore is
(75.8 g/(1714 g)) × 100 % = 4.4224 %.
(iv) The mole fraction can be found as
Mole fraction = (Number of moles of a component)/(Total number o moles in the solution)
Number of moles of toluene in the solution = 0.823 moles
Total number of moles in the solution = n_Toluene + n_benzene
= 0.823 moles + 1.224 moles = 2.05 moles
Mole fraction = 0.823 moles/(2.05 moles) = 0.402.