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Chemistry question

Dhamnekar Winod

Active member
Nov 17, 2018
149
Hello,


A rock salt (NaCl), ice, and water mixture is used to cool milk and cream to make homemade ice cream. How many grams of rock salt must be added to water to lower the freezing point 14.7°C? (Assume that there is 1 kg of water.)


How to answer this question? Answer provided is 231 grams. Is that correct?

If any member knows the correct answer to this question may reply.











 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
From wiki's Freezing-point depression:

If the solution is treated as an ideal solution, the extent of freezing-point depression depends only on the solute concentration that can be estimated by a simple linear relationship with the cryoscopic constant (Charles Blagden's Law):
$$ΔT_F = K_F \cdot b \cdot i,$$
where:
  • $ΔT_F$, the freezing-point depression.
  • $K_F$, the cryoscopic constant, which is dependent on the properties of the solvent, not the solute. For water, $K_F = 1.853\,K·kg/mol$.
  • $b$ is the molality (moles solute per kilogram of solvent).
  • $i$ is the Van 't Hoff factor (number of ion particles per individual molecule of solute, e.g. $i = 2$ for $\ce{NaCl}$, $3$ for $\ce{Ba Cl2}$).
 

Dhamnekar Winod

Active member
Nov 17, 2018
149
Hello,

There is error in your wikipedia link. $K_F$ shuld be $K_F=1.853^\circ C \cdot kg/mol$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
Hello,

There is error in your wikipedia link. $K_F$ shuld be $K_F=1.853^\circ C \cdot kg/mol$
It's the same thing.
The kelvin is the same unit as the degree Celsius except for the choice of its zero.
In this case we're talking about a change in temperature so that the zero of the scale does not matter.
 

Dhamnekar Winod

Active member
Nov 17, 2018
149
It's the same thing.
The kelvin is the same unit as the degree Celsius except for the choice of its zero.
In this case we're talking about a change in temperature so that the zero of the scale does not matter.
Hi,
I have answer to this question from another chemistry expert.

Molar mass of NaCl = 23.0 + 35.5 = 58.5 g/mol

van't Hoff factor for NaCl = 1 (1 mol NaCl = 1 mol Na⁺ + 1 mol Cl⁻)
Cryoscopic constant for water, $K_F = 1.853 ^\circ C\cdot kg/mol$
Freezing point depression, ΔTF = 14.7°C
Molality, b = ? mol/kg

$\Delta T_F = K_F\cdot b\cdot i$
14.7 = 1.853 × b × 2
Molality, b = 14.7 / (1.853 × 2) = 3.967 mol/kg water

(3.967 mol NaCl / 1 kg water) × (1 kg water) × (58.5 g NaCl / 1 mol NaCl)
= 232 g NaCl

In the above answer, $1.853^\circ C$ is used to arrive at the final answer 232 grams of NaCl. But you are saying it doesn't make any difference whichever units of temperature are used. How is that? In my opinion, we have to convert either degree Celsius into Kelvin or vice versa. Would you explain your standpoint in this regard?
 

Purplepixie

New member
Jul 11, 2020
4

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
In the above answer, $1.853^\circ C$ is used to arrive at the final answer 232 grams of NaCl. But you are saying it doesn't make any difference whichever units of temperature are used. How is that? In my opinion, we have to convert either degree Celsius into Kelvin or vice versa. Would you explain your standpoint in this regard?
Consider that we calculate a freezing point depression $\Delta T_F$, which is a change in temperature of the freezing point.
In $^\circ C$ the freezing point of water happens to be at $0\,^\circ C$ while in $K$ it is at $273.15\,K$.

If we calculate the freezing point depression with $^\circ C$ we find $\Delta T_F=14.7\,^\circ C$, which means that the freezing point of water is not at $0\,^\circ C$ but at $-14.7\,^\circ C$.
If instead we calculate with $K$ we find $\Delta T_F=14.7\,K$, which means that the freezing point of water is not at $273.15\,K$ but at $273.15\,K-14.7\, K$.
See how these statements are the same?