What percentage of NaCl in water will freeze at -10 C?

Question

As an Amazon Associate, I earn from qualifying purchases. For more practice problems like this, see Schaum’s Outline of College Chemistry, 10th Edition.

1. What the problem is really asking

We are asked: How much sodium‑chloride must be dissolved in water so that the solution begins to freeze at –10 °C?

In other words, we need the composition (mass % NaCl) of a water‑salt solution whose freezing point is –10 °C.
(The “percentage” can be expressed either as

* mass of salt ÷ mass of water (g NaCl / g H₂O)

or

* mass of salt ÷ total mass of solution (g NaCl / (g NaCl + g H₂O)).

Both are given, because they differ by a factor of (\frac{1}{1+\text{mass % of water}}).)

2. Step‑by‑step solution

2.1. Use the freezing‑point‑depression relation

For dilute electrolyte solutions the freezing point depression is well described by

[ \Delta T_f = i\,K_f\,m ]

where

(\Delta T_f) = depression of the freezing point (°C)
(i) = van’t Hoff factor (number of particles the solute produces). For NaCl, (i \approx 2) because it dissociates into Na⁺ and Cl⁻.
(K_f) = cryoscopic constant of water = 1.86 °C·kg mol⁻¹
(m) = molality of the solution (mol solute · kg⁻¹ solvent)

We want the solution to start freezing at (-10\;^{\circ}\text{C}). Pure water freezes at (0\;^{\circ}\text{C}), so

[ \Delta T_f = 0 - (-10) = 10\;^{\circ}\text{C} ]

Insert the numbers and solve for the required molality:

[ \begin{aligned} 10 &= (2)(1.86)\,m
m &= \frac{10}{2 \times 1.86} = \frac{10}{3.72} = 2.688\;\text{mol kg}^{-1} \end{aligned} ]

Thus 2.688 mol of NaCl per kilogram of water will depress the freezing point to –10 °C.

2.2. Convert molality to mass of NaCl

Molar mass of NaCl

[ M_{\text{NaCl}} = 22.99 + 35.45 = 58.44\;\text{g mol}^{-1} ]

Mass of NaCl needed per kilogram (1000 g) of water:

[ \begin{aligned} m_{\text{NaCl}} &= m \times M_{\text{NaCl}}
&= 2.688\;\frac{\text{mol}}{\text{kg}}\times 58.44\;\frac{\text{g}}{\text{mol}}
&= 157.2\;\text{g} \end{aligned} ]

So ≈ 157 g of NaCl per 1000 g of water gives a freezing point of –10 °C.

2.3. Express the result as a percentage

Definition	Calculation	Result
% ( NaCl / water ) (mass of salt ÷ mass of water)	(\displaystyle \frac{157.2\;\text{g}}{1000\;\text{g}}\times100)	15.7 % (w/w of water)
% ( NaCl / solution ) (mass of salt ÷ total mass)	(\displaystyle \frac{157.2}{1000+157.2}\times100)	13.6 % (w/w of solution)

Both numbers are correct; they just use different reference bases.
The value that is most often quoted in tables and handbooks is the solution‑mass percentage: ≈ 13–14 % NaCl by mass.

2.4. Why the simple equation works here

The linear freezing‑point‑depression equation is exact only for ideal dilute solutions.
For NaCl‑water, deviations become noticeable above ~20 % w/w. At ~14 % the error is < 1 °C, so the result above (≈ 13.5 % w/w) is accurate enough for practical purposes.

A more precise value can be read from an experimental phase diagram (e.g., the one you found). The diagram shows that a liquid of about 14 % NaCl (solution mass) has a liquidus line intersecting the temperature –10 °C, confirming our calculation.

3. Final answer

Mass of NaCl needed: ≈ 157 g per 1 kg (1000 g) of water.
Expressed as a percentage:
- 15.7 % NaCl / water (salt mass divided by water mass)
- 13.6 % NaCl / solution (salt mass divided by total mass)

Thus, a solution that is about 14 % NaCl by total mass (or ≈ 16 % relative to the water mass) will start to freeze at –10 °C.

4. Common Mistakes

Mistake	Why it’s wrong	How to avoid it
Using the simple formula and forgetting the van’t Hoff factor (i = 2 for NaCl)	Leads to a molality half the required value, giving a too‑low salt concentration.	Always include (i) for electrolytes; for NaCl, (i≈2).
Confusing “% w/w” with “% v/v”	Volume percentages are not appropriate for freezing‑point calculations, which depend on mass (or moles).	Use mass‑based percentages (mass of solute per mass of solvent or per total mass).
Reading the phase diagram incorrectly (e.g., taking the temperature axis as Celsius when it’s actually Fahrenheit)	Gives a completely wrong concentration.	Verify axis units; most scientific phase diagrams use °C.
Assuming the linear depression equation works at any concentration	At high concentrations the solution is non‑ideal; the linear relation under‑estimates the required salt.	For concentrations > 20 % w/w, consult experimental data or use activity‑coefficient models.
Reporting the percentage without specifying the reference (water vs. solution)	The answer can appear contradictory (13 % vs. 16 %).	State clearly whether the percentage is “% NaCl / water” or “% NaCl / solution”.

Keeping these points in mind will give a reliable estimate of the salt concentration needed for a desired freezing temperature.

Original question: What percentage of NaCl in water will freeze at -10 C? on Chemistry Stack Exchange, licensed CC BY-SA.