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

To calculate the mass of sodium chloride dissolved in a solution, we must use the boiling point elevation concept and the relationship ΔTb = Kb × m, which relates the boiling point elevation to the molality of the solution.

Step-by-step explanation:

The subject of this question is the calculation of the mass of sodium chloride dissolved in water using the concept of boiling point elevation. The boiling point of a solution increases when a solute, in this case sodium chloride (NaCl), is dissolved in a solvent, which is water. The boiling point elevation is directly proportional to the molality of the solution, and it is a colligative property. This means it depends only on the number of solute particles in the solution, not their identity.

The relationship between the boiling point elevation (ΔTb), the molal boiling-point elevation constant (Kb), and the molality (m) of the solution is expressed by the equation ΔTb = Kb × m. To find the mass of NaCl, we would first calculate the molality of the solution from the boiling point elevation, then use the molality along with the molar mass of NaCl to find the mass of NaCl dissolved.

Answer 2

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.

Step-by-step 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}} * \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 * K_(b) * \text{molality of solution}`

`(127.4 - 124.2)^(o)C = 2 * 0.62 * (m)/(60) * (1000)/(650)`

m = 100 g

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