Question

4.How many grams of NaCl would need to be added to 1001 g of water to increase the boiling temperature of the solution by 1.500 °C? (Kb for water is 0.5100 °C/m)

5. A solution is made using 80.1 g of toluene (MM = 92.13 g/mol) and 80.0 g of benzene (MM = 78.11 g/mol). What is the molality of the toluene in the solution?

Answer 1

(4) The mass of NaCl needed would be, 172.2 grams.

(5) The molality of toluene in the solution is, 10.9 mol/kg

Explanation :

Part 4:

`(K_b)` for water =
`0.5100^oC/m`

`\Delta T_b=1.500^oC`

Mass of water (solvent) = 1001 g = 1.001 kg

Molar mass of NaCl = 58.5 g/mole

Formula used :

`\Delta T_b=i* K_b* m\\\\\Delta T_b=i* K_b*\frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}* \text{Mass of water in Kg}}`

where,

`\Delta T_b` = change in boiling point

i = Van't Hoff factor = 2 (for NaCl electrolyte)

`K_b` = boiling point constant for water

m = molality

Now put all the given values in this formula, we get

`1.500^oC=2* (0.5100^oC/m)* \frac{\text{Mass of NaCl}}{58.5g/mol* 1.001kg}`

`\text{Mass of NaCl}=172.2g`

Therefore, the mass of NaCl needed would be, 172.2 grams.

Part 5:

Formula used :

`\text{Molality}=\frac{\text{Mass of toluene}* 1000}{\text{Molar mass of toluene}* \text{Mass of benzene (in g)}}`

Given:

Mass of toluene = 80.1 g

Mass of benzene = 80.0 g

Molar mass of toluene = 92.13 g/mol

`\text{Molality}=(80.1g* 1000)/(92.13g/mole* 80.0g)=10.9mole/kg`

Therefore, the molality of toluene in the solution is, 10.9 mol/kg