Question

A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving some sodium chloride () in of . This solution boils at . Calculate the mass of that was dissolved. Round your answer to significant digits.

Answer 1

The given question is incomplete. The complete question is as follows.

A certain liquid has a normal boiling point of $124.2^{o}C$ and a boiling point elevation constant $k_{b} = 0.62 ^{o}C kg mol^{-1}$. A solution is prepared by dissolving some sodium chloride (NaCl) in 6.50 g of X. This solution boils at $127.4^{o}C$. 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 =

$T_{o}$ = boiling point of the pure solvent = $124.2^{o}C$

$K_{b}$ = elevation of boiling constant = $0.62 ^{o}C kg mol^{-1}$

We will calculate the molality as follows.

     molality = $\frac{\text{maas of solute}}{\text{molar mass of solute}} \times \frac{1000}{\text{mass of solvent in g}}$

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.  

    $T - T_{o} = i \times K_{b} \times \text{molality of solution}$

  $(127.4 - 124.2)^{o}C = 2 \times 0.62 \times \frac{m}{60} \times \frac{1000}{650}$

                  m = 100 g

Therefore, we can conclude that 100 g of NaCl was dissolved.