What is the freezing point of a 2.12 molal solution of NaCl in water if the Kf is -1.86 ?

Question

UpStudy AI · Accepted Answer

To determine the freezing point of a 2.12 molal (m) solution of NaCl in water, we use the freezing point depression formula:

$$
\Delta T_f = i \times K_f \times m
$$

Where:
- $$\Delta T_f$$ = change in freezing point
- $$i$$ = van 't Hoff factor (number of particles the solute dissociates into)
- $$K_f$$ = cryoscopic constant (freezing point depression constant)
- $$m$$ = molality of the solution

**Given:**
- $$m = 2.12$$ mol/kg
- $$K_f = 1.86$$ °C·kg/mol (Note: Freezing point depression constants are typically positive, even if the question provides a negative value, because the formula accounts for the depression itself.)
- For NaCl, $$i = 2$$ (since NaCl dissociates into Na$$^+$$ and Cl$$^-$$)

**Calculation:**

$$
\Delta T_f = 2 \times 1.86 \times 2.12 = 7.89 \text{ °C}
$$

This means the freezing point is lowered by 7.89°C from that of pure water.

**Final Freezing Point:**

$$
\text{Freezing point} = 0°C - 7.89°C = -7.89°C
$$

**Rounded to one decimal place:**

$$
\text{Freezing point} \approx -7.9°C
$$

**Answer:**  
The solution freezes at approximately –7.9 °C.
The freezing point of the 2.12 molal NaCl solution is approximately –7.9 °C.

What is the freezing point of a 2.12 molal solution of NaCl in water if the Kf is -1.86 ?

Real Tutor Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Chemistry Questions