What will be the boiling point of a solution of 8 moles of sodium dichromate (Na2Cr2O7) dissolved in 8 kg of water? Use the following values: Kb = 0.512 K · m−1 Kf = 1.86 K · m−1 1. 374.69 K 2. 373.92 K 3. 378.73 K 4. 380.27 K 5. 377.76 K

1. 374.69 K

Explanation:

Hello,

In this case, pure water's boiling point is 373.15 K, thus by considering the boiling point increase equation:

Whereas i=2 since two ionic species are formed,actually, the experimental value is 2.42 so better work with it, thus:

Thus, the required boiling point is:

Regards.

374.39 K

Explanation:

First, we will calculate the molality of the solution.

b = 8 mol / 8 kg = 1 m

We can find the elevation in the boiling point using the following expression.

ΔTb = i × Kb × b

ΔTb = i × Kb × b

ΔTb = 2.42 × 0.512 K × m⁻¹ × 1 m

ΔTb = 1.24 K

The normal boiling point of water is 373.15 K. The boiling point of the solution is:

373.15 K + 1.24 K = 374.39 K

the volume increases