A 7.27-gram sample of a compound is dissolved in 250 grams of benzene. The freezing point of this solution is 1.02°C below that of pure benzene. What is the molar mass of this compound? (Note: Kf for benzene = 5.12°C/m.) Ignore significant figures for this problem:. a. 36.5 g/mol
b. 146 g/mol
c. 292 g/mol
d. 5.79 g/mol
e. 73.0 g/mol

146 g/mol → option b.

Explanation:

This is a problem about the freezing point depression. The formula for this colligative property is:

ΔT = Kf . m . i

We assume i = 1, so our compound is not electrolytic.

ΔT = Freezing T° of pure solvent - Freezing T° of solution = 1.02 °C

m  = molality (mol of solute/kg of solvent)

We convert the grams of solvent (benzene) to kg → 250 g . 1 kg/1000 = 0.250 kg.

We replace → 1.02°C = 5.12°C/mol/kg . mol/ 0.250kg . 1

1.02°C / 5.12 mol/kg/°C = mol/ 0.250kg

0.19922 mol/kg = mol/ 0.250kg

mol = 0.19922 . 0.250kg → 0.0498 mol

molar mass = g/mol → 7.27 g / 0.0498mol = 146 g/mol

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be

c-14 is carbon-14, or radiocarbon.   it is a radioactive isotope of carbon with an atomic nucleus containing 6 protons and 8 neutrons.