Respuesta :
Answer :
(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:
[tex](K_b)[/tex] for water = [tex]0.5100^oC/m[/tex]
[tex]\Delta T_b=1.500^oC[/tex]
Mass of water (solvent) = 1001 g = 1.001 kg
Molar mass of NaCl = 58.5 g/mole
Formula used :
[tex]\Delta T_b=i\times K_b\times m\\\\\Delta T_b=i\times K_b\times\frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}\times \text{Mass of water in Kg}}[/tex]
where,
[tex]\Delta T_b[/tex] = change in boiling point
i = Van't Hoff factor = 2 (for NaCl electrolyte)
[tex]K_b[/tex] = boiling point constant for water
m = molality
Now put all the given values in this formula, we get
[tex]1.500^oC=2\times (0.5100^oC/m)\times \frac{\text{Mass of NaCl}}{58.5g/mol\times 1.001kg}[/tex]
[tex]\text{Mass of NaCl}=172.2g[/tex]
Therefore, the mass of NaCl needed would be, 172.2 grams.
Part 5:
Formula used :
[tex]\text{Molality}=\frac{\text{Mass of toluene}\times 1000}{\text{Molar mass of toluene}\times \text{Mass of benzene (in g)}}[/tex]
Given:
Mass of toluene = 80.1 g
Mass of benzene = 80.0 g
Molar mass of toluene = 92.13 g/mol
[tex]\text{Molality}=\frac{80.1g\times 1000}{92.13g/mole\times 80.0g}=10.9mole/kg[/tex]
Therefore, the molality of toluene in the solution is, 10.9 mol/kg
