[Atome]

Hello there:

I am having some trouble with the following problem and would appreciate help. Thank you!

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1. Why is road salt not used in cold areas during winter? Back up your answer with a calculator. (Helpful info: The maximum solubility of NaCl in water is ca. 37 g NaCl/100 g H₂O at 0°C)

 :delta:T_fr = K_f x C_m x i

Molality = mol solute/kg solvent = 0.6330 mol NaCl/[(100 - 37) g x 0.001 kg/g] = 10.05 molal

Therefore:

T_fr = 1.858 °C/molal x 10.05 molal x 2 = 37°C.

I am unsure about this answer because it is awfully high; the freezing point of water would then be -37°C.

[enahs]

Check the cancellation of your units when you calculated the molality.

[Atome]

Thanks for your reply, enahs.

However, I am not seeing any mistakes in my algebra with the units:

Molality = mol solute/kg solvent = 0.6330 mol NaCl/[(100 - 37) g x 0.001 kg/g] = 10.05 molal

= mol ÷ ( g x kg/g) = mol ÷ kg = molal

Am I right?

[enahs]

Write it out on paper if you have to.

You have it as:
0.6330 mol NaCl   *    0.001kg
(100 - 37)g                1g

Do your units cancel?

[enahs]

I am sorry, I did not notice your square bracket. You are correct.

Your 100-37 is a problem though.
If I add 37 g of NaCl to 100 grams of water, did the mass of the water change?

[Atome]

Thank you very much for your help again, enahs.

I believe that I have the correct answer. I originally thought that 100 g was the solution and not the solvent.

For the freezing point depression, I got 24°C. Since areas with cold climates can reach temperatures below -24°C, road salt would not be effective in inhibiting ice formation in such frigid regions.