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

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 =

The depression in freezing point can be given as:

Depression in freezing point = Van't Hoff factor × Freezing point constant ×   ......(i)

Given, the depression in freezing point = 2.74

Van't Hoff factor = 1 (Molecular compound)

Freezing point constant (Kf) = 5.12 /m

Mass of solute = 26.7 g

Mass of solvent = 0.250 kg

Substituting the values in equation (i):

2.74 = 1 × 5.12

=

0.535 =

Molecular mass of solute = 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:

link

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

Explanation:

Depression in freezing point is given by:

= Depression in freezing point

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

= freezing point constant =

m= molality

Weight of solvent (benzene)= 0.250 kg  

Molar mass of solute = M g/mol

Mass of solute  = 26.7 g

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

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.

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