Respuesta :
The question is: If 40 g of a compound C6H10O5 are dissolved in 500 g of water, determine the boiling point of this solution. (Water: boiling temperature 100 ° C and Kb = 0.52 ° C / m).
Answer: The boiling point of given solution is [tex]100.256^{o}C[/tex].
Explanation:
Given: Mass of solute = 40 g
Mass of solvent = 500 g (1 g = 0.001 kg) = 0.5 kg
[tex]K_{b} = 0.52^{o}C/m[/tex]
Molality is the moles of solute present in kg of solvent.
Moles is the mass of substance divided by its molar mass. Hence, moles of [tex]C_{6}H_{10}O_{5}[/tex] (molar mass = 162.141 g/mol) is calculated as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{40 g}{162.141 g/mol}\\= 0.246 mol[/tex]
Now, molality of the solution is calculated as follows.
[tex]Molality = \frac{molesof solute}{mass(in kg)}\\= \frac{0.246}{0.5 kg}\\= 0.492 m[/tex]
The boiling point is calculated as follows.
[tex]\Delta T_{b} = K_{b}m[/tex]
where,
m = molality
Substitute the values into above formula as follows.
[tex]\Delta T_{b} = K_{b}m\\= 0.52^{o}C/m \times 0.492 m\\= 0.256^{o}C[/tex]
As the boiling point of water is 100 degree Celsius. So, the boiling point of solution is as follows.
[tex](100 + 0.256)^{o}C\\= 100.256^{o}C[/tex]
Thus, we can conclude that the boiling point of given solution is [tex]100.256^{o}C[/tex].
