Thermal Properties Of Matter

The value represents the ratio of the distance travelled by the substance to the distance travelled by the solvent front. In this case, the distance for is 8 cm and for the solvent front is 12 cm. Hence, is calculated as: Thus, the correct value for of B is .

We are tasked with finding the atomic masses of and based on the freezing point depression ( ) of solutions of and in benzene ( ). The given data is:

Mass of ,
Mass of benzene = 20 g,
for solution = 2.318 K,
Mass of ,
for solution = 1.314 K,
of benzene = 5.1 K kg mol .

Step 1: Freezing point depression formula

The freezing point depression is given by:

where:

is the cryoscopic constant,

is the molality of the solution.

The molality is given by:

Step 2: Calculate molality for

For :

The mass of benzene is 20 g = 0.02 kg. Thus, the moles of are:

The molar mass of is:

Step 3: Calculate molality for

For :

The mass of benzene is 20 g = 0.02 kg. Thus, the moles of are:

The molar mass of is:

Step 4: Set up equations for atomic masses

Let the atomic mass of be and the atomic mass of be .

For :

For :

Step 5: Solve the equations

Subtract the first equation from the second:

Substituting into the first equation:

Step 6: Match with the options

The atomic masses are and , which matches option (4).

Final Answer: