Chemistry, 16.11.2019 06:31 brandonleestewovilgw

What is the molar mass of toluene if 0.85 g of toluene decreases the freezing point of 100. g of benzene by 0.47°c? kf of benzene is 5.12°c/m.

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Answer from: 2020davidhines

- Molar mass is 92.59 gram per mol.

Solution:- Depression in freezing point is directly proportional to the molality of the solution. The equation we use for this type of problems is:

$\Delta T_f=imk_f$

where, $\Delta T_f$ is depression in freezing point, i is Van't hoff factor, m is molality and kf is the freezing point depression constant.

Toluene is a non electrolyte and so the value of i for this is 1.

depression in freezing point is given as 0.47 and the kf is given as 5.12. So, we could calculate the molality of the solution using the equation written above.

Let's plug in the values in the equation:

$0.47^0C=1*m*5.12\frac{^0C}{m}$

$m=\frac{0.47^0C}{5.12\frac{^0C}{m}}$

m = 0.0918 m

It means the molality of the solution is 0.0918 mol/kg.

We know that molality is, moles of solute per kg of the solvent. Mass of solvent is 100. g that is 0.100 kg. Now, we could calculate the moles of the solute that is toluene as:

$molality=\frac{mol}{kg}$

$0.0918=\frac{mol}{0.100}$

mol=0.0918*0.100

mol = 0.00918

Mass of solute is given as 0.85 g. We divide the grams by the moles to get the molar mass.

Molar mass = $\frac{0.85g}{0.00918mol}$

Molar mass = $\frac{92.59g}{mol}$

So, the molar mass of toluene is 92.59 gram per mol.

Answer from: dean111375

92.59 g/mol is the molar mass of toluene.

Explanation:

Freezing point of pure benzene= T =

Freezing point of sample = T_f =

Depression in freezing point = $\Delta T_f=0.47^oC$

Depression in freezing point is also given by formula:

$\Delta T_f=i\times K_f\times m$

K_f = The freezing point depression constant

m = molality of the sample

i = van't Hoff factor

We have: K_f = 5.12°C kg/mol

i = 1 ( organic compounds)

$\Delta T_f=14^oC$

$0.47^oC=1\times 5.12^oC kg/mol\times m$

$m=\frac{0.47^oC}{1\times 5.12^oC kg/mol}=0.0918 mol/kg$

$molality(m)=\frac{moles(n)}{\text{mass of solvent(kg)}}$

$0.0918=\frac{n}{0.1 kg}$

n = 0.00918 mol

$n=\frac{\text{Mass of compound}}{\text{molar mass of compound}}$

$0.00918 mol=\frac{0.85 g}{\text{molar mass of compound}}$

Molar mass of compound(toluene) :

$\frac{0.85 g}{0.00918 mol}=92.59 g/mol$

92.59 g/mol is the molar mass of toluene.

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What is the molar mass of toluene if 0.85 g of toluene decreases the freezing point of 100. g of ben...