Calculate the freezing point of a solution made from 22.0 g of octane (c8h18) dissolved in 148.0 g of benzene. benzene freezes at 5.50°c and its kf value is 5.12°c/m. a.–1.16°c b.0.98°c c.6.66°c d.12.2°c e.5.49°c

The equation that represents the relation between freezing point depression and molality is,

Δ

Where,

Δ is the change in the freezing point of the solvent.

of benzene is 5.12

m is molality of the solution

Moles of solute = = 0.193 mol

Mass of solvent in kg =

Molality =

= 1.30 m

Δ

Δ ) (0.130 m) " alt=" 5.12^{0}C/m" />) (0.130 m) " />

= 6.66

Δ

So, the correct answer is a. -1.16

the volume increases