Question

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

Answer 1

Answer:

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:

$\Delta T=imKf$

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

$\Delta T=2.42*(\frac{8}{8}m)*0.512 Km^{-1}=1.239K$

Thus, the required boiling point is:

$T_b=373.15K+1.239K\\T_b=374.69K$

Regards.

Answer 2

Answer:

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

• i:  van 't Hoff factor (experimental value 2.42)
• Kb: ebullioscopic constant

• b: molality

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