Suppose that you add 26.7 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 2.74 oC compared to pure benzene. What is the molar mass of the unknown compound

Respuesta :

Answer: The molar mass of the unknown compound is 200 g/mol

Explanation:

Depression in freezing point is given by:

[tex]\Delta T_f=i\times K_f\times m[/tex]

[tex]\Delta T_f=2.74^0C[/tex] = Depression in freezing point

i= vant hoff factor = 1 (for molecular compound)

[tex]K_f[/tex] = freezing point constant = [tex]5.12^0C/m[/tex]

m= molality

[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]

Weight of solvent (benzene)= 0.250 kg  

Molar mass of solute = M g/mol

Mass of solute  = 26.7 g

[tex]2.74^0C=1\times 5.12\times \frac{26.7g}{Mg/mol\times 0.250kg}[/tex]

[tex]M=200g/mol[/tex]

Thus the molar mass of the unknown compound is 200 g/mol

The molar mass of an unknown solute compound in the solution has been 199.626 g/mol.

With the addition of the solute to the solution, there has been a depression in the freezing point. The depression in the freezing point can be expressed as:

Depression in freezing point = Van't Hoff factor × Freezing point constant × molality

The molality can be defined as the moles of solute per kg solvent

Molaity = [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{1}{Mass\;of\;solvent\;(kg)}[/tex]

The depression in freezing point can be given as:

Depression in freezing point = Van't Hoff factor × Freezing point constant ×  [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{1}{Mass\;of\;solvent\;(kg)}[/tex] ......(i)

Given, the depression in freezing point = 2.74 [tex]\rm ^\circ C[/tex]

Van't Hoff factor = 1 (Molecular compound)

Freezing point constant (Kf) = 5.12 [tex]\rm ^\circ C[/tex]/m

Mass of solute = 26.7 g

Mass of solvent = 0.250 kg

Substituting the values in equation (i):

2.74 [tex]\rm ^\circ C[/tex] = 1 × 5.12

[tex]\rm \dfrac{2.74}{5.12}[/tex] = [tex]\rm \dfrac{1}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{26.7}{0.250\;kg}[/tex]

0.535 = [tex]\rm \dfrac{1}{Molecular\;mass\;of\;solute}\;\times\;106.8[/tex]

Molecular mass of solute = [tex]\rm \dfrac{106.8}{0.535}[/tex] g/mol

Molecular mass of solute = 199.626 g/mol

The molar mass of an unknown solute compound in the solution has been 199.626 g/mol.

For more information about the molality of the compound, refer to the link:

https://brainly.com/question/7229194