The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of benzene (Kf of benzene = 4.90°C/m)?

The freezing point of solution is 5.35°C

Explanation:

The equation used to calculate depression in freezing point follows:

To calculate the depression in freezing point, we use the equation:

Or,

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

= molal freezing point elevation constant = 4.90°C/m

= Given mass of solute (naphthalene) = 2.60 g

= Molar mass of solute (naphthalene) = 128.2 g/mol

= Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

Hence, the freezing point of solution is 5.35°C

if the concentration of products is decreased, the equilibrium will shift toward the products). add an inert gas (one that is not involved in the reaction) to the constant-volume reaction mixture: this will increase the total pressure of the system, but will have no effect on the equilibrium condition.

explanation:

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm