Question

What is the molar mass of a solution of 5.00 g of a compound in 25.00 g of carbon tetrachloride (bp 76.8 °c; kb = 5.02 °c/m) that boils at 81.5 °c at 1 atm? (a) outline the steps necessary to answer the question?

Answer 1

(84.4C - 76.5C) / (5.03 C/m) = 1.5706 m

(1.5706 mol) / (1000 g CC14) X (25.00 G CC14) = 0.039265 mol

(5.00 g) / (0.039265 mol) = 127 g/mol

Answer 2

Answer: 214 g/mol

Step-by-step explanation:

Formula used for Elevation in boiling point :

`\Delta T_b=k_b* m`

or,

`T_(solution)-T_(solvent)=k_b* \frac{\text{Mass of solute}}{\text {Molar mass of solute}* \text{ Mass of solvent in Kg}}`

where,

`T_b` = change in boiling point =
`(81.5-76.8)^0C=4.7^0C`

`k_b` = boiling point constant =
`5.02^0C/m`

m = molality =
`\frac{5g}{\text{ Molar mass of solute}}* 0.025kg`

`4.7=5.02* \frac{5g}{\text {Molar mass of solute}* 0.025kg}`

`{\text {Molar mass of solute}}=214g/mol`

Thus molar mass of solute is 214 g/mol.