Respuesta :
The formula used to determine the molarity is:
[tex]molarity = \frac{number of moles of solute}{volume of solution in Liter}[/tex] -(1)
Molality = [tex]5.39 m[/tex] (given)
As, molality = [tex]\frac{number of moles of solute}{mass of solvent in kg}[/tex]
That means 5.39 moles of solute are present in 1000 g of solvent.
Moles of solute, benzene = 5.39 moles.
Molar mass of benzene, [tex]C_6H_6 = 6\times 12+6\times 1 = 78 g/mol[/tex]
Using, [tex]number of moles = \frac{mass of compound}{Molar mass of compound}[/tex] we can determine the mass of benzene as:
[tex]mass of benzene = number of moles of benzene \times Molar mass of benzene[/tex]
[tex]mass of benzene = 5.39 mol\times 78 g/mol = 420.42 g[/tex]
Mass of solvent = 1000 g
Mass of solution = mass of solute + mass of solvent
Mass of solution = [tex]1000 g + 420.42 g = 1420.42 g[/tex]
Density of solution = [tex]0.869 g/mL[/tex] (given)
Since, [tex]density = \frac{mass}{volume}[/tex]
So, Volume of solution = [tex]\frac{1420.42 g}{0.869 g/mL}[/tex] = [tex]1634.54 mL[/tex]
Since, [tex]1 mL = 0.001 L[/tex]
So, volume of solution = [tex]1.634 L[/tex]
Substituting the values in formula (1):
[tex]molarity = \frac{5.39 mol}{1.634 L} = 3.29 mol/L[/tex]
Hence, the molarity of the solution is [tex]3.29 M[/tex].
