Cyclohexane has a freezing point of 6.50 ∘c and a kf of 20.0 ∘c/m. what is the freezing point of a solution made by dissolving 0.617 g of biphenyl (c12h10) in 25.0 g of cyclohexane? express the temperature numerically in degrees celsius.

Temperature of freezing for the solution is. 3.3°C

Explanation:

This is an excersise of coligative property (freezing low depression)

ΔT = Kf . m

Mass of solute (biphenyl) → 0.617 g

Mass of solvent (cycloexane) → 25 g

m = molality (mol of solute in 1kg of solvent)

Let's find out moles of solute:

Moles = Mass / Molar mass

Moles = 0.617 g / 154 g/m → 0.004 moles

To know molalilty, let's convert 25 g to kg

25 g = 0.025 kg

moles / kg = molality

0.004 m / 0.025 kg = 0.16m

ΔT = T°freezing pure solvent  - T° freezing solution

6.5°C - T° freezing solution  = 20°C/m . 0.16m

6.5°C - T° freezing solution = 3.20°C

6.5°C - 3.20°C = T° freezing solution → °3.3 °C

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.

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