A 0.2436-g sample of an unknown substance was dissolved in 20.0 ml of cyclohexane. The density of cyclohexane is 0.779 g/ml. The freezing-point depression was 2.5ºc. Calculate the molar mass of the unknown substance

Respuesta :

The depression in freezing point is related to molality of solution as follows:

[tex]\Delta T_{f}=k_{f}m[/tex]

Here, [tex]k_{f}[/tex] is freezing point depression constant, for cyclohexane it is equal to 20°C kg/mol.

The value of freezing-point depression is 2.5 °C, molality can be calculated as follows:

[tex]m=\frac{\Delta T_{f}}{k_{f}}=\frac{2.5 ^{o}C}{20 ^{o}C kg/mol}=0.125 mol/kg[/tex]

Molality is defined as number of moles of solute in 1 kg of solvent.

Here, solvent is cyclohexane, its volume is given 20.0 mL and density is 0.779 g/mol thus, mass of cyclohexane can be calculated as follows:

[tex]m=d\times V=0.779 g/mL\times 20 mL=15.58 g[/tex]

Converting this into kg,

[tex]1 g=10^{-3} kg[/tex]

Thus, 15.58 g will be 0.01558 kg.

Now, number of moles of unknown solute is related to its mass and molar mass as follows:

[tex]n=\frac{m}{M}[/tex]

Putting the values of mass of solute which is 0.2436 or 0.0002436 kg

[tex]n=\frac{ 0.0002436 kg}{M}[/tex]

Now, expression for molality of solution is:

[tex]m=\frac{n_{solute}}{m_{solvent}}[/tex]

Putting all the values,

[tex]0.125 mol/kg=\frac{0.0002436 kg}{M\times 0.01558 kg}[/tex]

Or,

[tex]0.125 mol/kg=\frac{0.015635}{M}[/tex]

On rearranging,

[tex]M=\frac{0.015635}{0.125 mol/kg}=0.1250 kg/mol[/tex]

Or,

[tex]M=0.125 kg/mol(\frac{1000 g}{1 kg})=125 g/mol[/tex]

Therefore, molar mass of unknown sample is 125 g/mol

The molar mass of the unknown substance is 125 g/mol.

What is cyclohexane?

Cyclohexane is a colorless, flammable liquid which has a detergent like odor.

It is a cycloalkane whose formula i s [tex]\bold{C_6H_1_2}[/tex].

Given,

Density is 0.779 g/ml

The freezing-point depression is 2.5ºc.

Mass is 0.2436-g

In kg = 0.0002436 kg

Now,

The depression in freezing point is equal to the molality of solution which is

[tex]\bold{\Delta T_f= k_fm}[/tex]

where [tex]\bold{ k_f}[/tex] is equals to freezing point depression constant, which is equal to 20.0 ml.

Step 1: Calculating the molality

[tex]\bold{m= \dfrac{\Delta T_f}{k_f} = \dfrac{2.5^\circ c}{20^\circ c\;kg/mol}=0.125\;mol/kg }[/tex]

Step 2: Calculating the Mass of cyclohexane

[tex]\bold{m= d\times V= 0.779\;g/ml \times20\;ml =15.58\;g}[/tex]

Converting mass into kg

[tex]\bold{1\;g= 10^-^3 kg}[/tex]

15.58 g = 0.01558 kg

Step 3: Calculating the number of moles

[tex]\bold{n = \dfrac{mass}{molar\;mass} = \dfrac{0.0002436\;kg}{M} }[/tex]

Step 4: Expression for the molality of the solution

[tex]\begin{aligned}\bold{m &= \dfrac{n_s_o_l_u_t_e}{m_s_o_l_u_t_e} \\\\0.125\;mol/kg & = \dfrac{0.015635}{M} \\\\M &= \dfrac{0.015635}{0.125\;mol/kg} = 0.1250\;kg/mol\\\\M &= 0.125\;kg/mol \dfrac{1000\;g}{1\;kg} = 125\;g/mol }\end{aligned}[/tex]

Thus, the molar mass of the substance is 125 g/ mol.

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