A studnet was required to make a solution of camphor and napthlaene that had a freezing point of 166C. How many grams of naphthalene need to be added to 10 g of camphor to make this oslution the freezing point of pure camphor is 178.4°C and the freezing point depression constant for camphor kf is 38.8°C/m.

205.9 grams of naphthalene

Explanation:

∆T = m × Kf

m (molality of the solution) = ∆T/Kf = (178.4°C - 166°C)/38.8°C/m = 12.4°C/38.8°C/m = 0.3196 m = 0.3196 mol/kg

Molality of solution = moles of camphor/mass of naphthalene in kilogram

Moles of camphor = 10/152 = 0.0658 mol

Mass of naphthalene = 0.0658/0.3196 = 0.2059 kg = 0.2059×1000 g = 205.9 g

im not exactly sure but i think its b!

i don't see a text, but i am gonna assume this *orbit* the curved path, usually elliptical, described by a planet, satellite, spaceship, etc., around a celestial body, as the sun.