Respuesta :
The given question is incomplete. The complete question is as follows.
A certain liquid has a normal boiling point of [tex]124.2^{o}C[/tex] and a boiling point elevation constant [tex]k_{b} = 0.62 ^{o}C kg mol^{-1}[/tex]. A solution is prepared by dissolving some sodium chloride (NaCl) in 6.50 g of X. This solution boils at [tex]127.4^{o}C[/tex]. Calculate the mass of NaCl that was dissolved. Round your answer to significant digits.
Explanation:
As per the colligative property, the elevation in boiling point will be as follows.
T = boiling point of the solution =
[tex]T_{o}[/tex] = boiling point of the pure solvent = [tex]124.2^{o}C[/tex]
[tex]K_{b}[/tex] = elevation of boiling constant = [tex]0.62 ^{o}C kg mol^{-1}[/tex]
We will calculate the molality as follows.
molality = [tex]\frac{\text{maas of solute}}{\text{molar mass of solute}} \times \frac{1000}{\text{mass of solvent in g}}[/tex]
i = vant hoff's factor
As NaCl is soluble in water and dissociates into sodium and chlorine ions so i = 2.
Putting the given values into the above formula as follows.
[tex]T - T_{o} = i \times K_{b} \times \text{molality of solution}[/tex]
[tex](127.4 - 124.2)^{o}C = 2 \times 0.62 \times \frac{m}{60} \times \frac{1000}{650}[/tex]
m = 100 g
Therefore, we can conclude that 100 g of NaCl was dissolved.
