A solution of 5.0 g benzoic acid (C6H5COOH) in 100.0 g of carbon tetrachloride has a boiling point of 77.5° C. The boling point of pure CCl4 is 76.5° C. Calculate the molar mass of benzoic acid in the solution based on the boiling point elevation. (Kb for CCl4 is 5.03° C/m.)

The molar mass of benzoic acid is 251.3 g/mol

Explanation:

∆T = m × Kb

m (molality) = ∆T/Kb = (77.5 - 76.5)/5.03 = 1/5.03 = 0.199 mol/kg

Molar mass of benzoic acid = mass of benzoic acid ÷ (molality × mass of CCl4 in kilograms) = 5 ÷ (0.199 × 100/1000) = 5 ÷ 0.0199 = 251.3 g/mol

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2