Yes, that is close to what I had on mind, just my approach took using Claussius-Clapeyron equation for sublimation and evaporation and an additional assumption (quite easy to defend). Had to solve it again to close the case

From the information given we know at the melting point pressure of the vapor is 611 Pa. What we also know (not from the question itself, but I assume it to be a common knowledge), is that the melting point (0 °C) is very close to the triple point (0.01 °C apart) - that means we can safely assume 611 Pa to be also a saturated vapor over the solid.

Next things were already said earlier, so I am just listing them here for completeness. At the melting point saturated vapor pressures over the solid and the liquid are identical:

[tex]p_{liquid} = p_{solid}[/tex]

we also know that the

[tex]p_{liquid} = (1-x)p_{liquid}^0[/tex]

and finally we have two Claussius-Clapeyron equations ([itex]t[/itex] stands for the melting point temperature):

[tex]\log\left(\frac{p_{solid}}{611 Pa}\right) = \frac {\Delta H_{sub}}R\left(\frac 1 {273.16} - \frac 1 t\right)[/tex]

and

[tex]\log\left(\frac{p_{liquid}^0}{611 Pa}\right) = \frac {\Delta H_{vap}}R\left(\frac 1 {273.16} - \frac 1 t\right)[/tex]

plugging [itex]p_{liquid}^0[/itex] into the second equation

[tex]\log\left(\frac{\frac{p_{liquid}}{1-x}}{611 Pa}\right) = \frac {\Delta H_{vap}}R\left(\frac 1 {273.16} - \frac 1 t\right)[/tex]

We have two equations, easy to solve for [itex]t[/itex] and [itex]p_{liquid}[/itex]. I was too lazy to do that manually, so I just plugged them into

Wolfram Alpha, and got t=271.33.

Again, solution is 1 molal, so ΔT=273.16-171.33=1.83 K×kg/mol.