Question

Use the result of the previous problem to calculate the freezing temperature of seawater.

Short answer

Therefore, the freezing temperature of seawater is$-2.167°\mathrm{C}$

Step-by-step solution

Given information

From the previous problem we have:

$T-T_0=-\frac{N_{B}k{T}_0^2}{L}$

Explanation

from the previous problem, we have

$T-T_0=-\frac{N_{B}k{T}_0^2}{L}$

but,$N_{B}k=n_B{R}_r$, so

$T-T_0=-\frac{n_{B}R{T}_0^2}{L}$

Where,

$T_o$is the freezing temperature of water

$n_B$is the number of moles of salt

$L$is the fusion latent energy of water and is given as $L=333.7\times {10}^{3}J$

Calculations

We have 35 grammes of salt in one kilogram of water because sea water contains calcium and sodium, which have an average molar mass of 30 g/mol, hence in 35g the number of moles are

$n_B=\frac{35\mathrm{g}}{30\mathrm{g}/\mathrm{mol}}=1.167\mathrm{mol}$

Substitute the values,

role="math" localid="1647204325114" $$\begin{gathered} T-T_0=-\frac{(1.167\mathrm{mol})(8.314\mathrm{J}/\mathrm{mol}·\mathrm{K})(273\mathrm{K}{)}^2}{333.7\times {10}^3\mathrm{J}} \\ =-2.167\mathrm{K} \\ \mathrm{T}=273\mathrm{K}-2.167\mathrm{K}=270.833\mathrm{K} \\ \mathrm{T}=-2.167°\mathrm{C} \end{gathered}$$