# A solution is made by dissolving 25.0 g of magnesium chloride crystals in 1000 g of water. What will be the freezing point of the new solution assuming complete dissociation of the #MgCl_2# salt?

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What will the boiling point of the new solution assuming complete dissociation of the #MgCl_2# salt?

What will the boiling point of the new solution assuming complete dissociation of the

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It can be stated that for every mole of magnesium chloride, three moles of ions will be produced in solution if the salt dissociates completely.

The freezing-point depression of a solution can be expressed mathematically using the following equation:

- Tcriosebu.pdf https://tutor.hix.ai

The ratio between the concentration of particles created in solution and the concentration of the dissolved material is known as the van der Hoff factor.

Use the compound's molar mass to accomplish that.

Molality is now defined as the number of solute moles in one kilogram of solvent.

In your situation, you'll have

Accordingly, the freezing-point depression is expected to be

The definition of the freezing-point depression is

This indicates that you've

Since you have one sig fig for the mass of water, you should round this to that number, but I'll leave it rounded to two sig figs.

This brings us to the boiling point of this solution: the boiling-point elevation equation looks like this.

- Tcriosebu.pdf https://tutor.hix.ai

Enter your values to obtain

The definition of the boiling-point elevation is

As for you, you've

This response will remain as

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To find the freezing point of the solution, we can use the formula:

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

Where:

- (\Delta T_f) is the freezing point depression,
- (i) is the van't Hoff factor (the number of particles produced when the substance dissociates),
- (K_f) is the cryoscopic constant for water (1.86 °C kg/mol), and
- (m) is the molality of the solution (moles of solute per kilogram of solvent).

First, we need to find the number of moles of magnesium chloride (MgCl2) in the solution:

- The molar mass of MgCl2 is approximately 95.21 g/mol.
- So, (25.0 , \text{g}) of MgCl2 is equal to (\frac{25.0 , \text{g}}{95.21 , \text{g/mol}}) = (\approx 0.263 , \text{mol}).

Next, we calculate the molality of the solution:

- (m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}})
- (m = \frac{0.263 , \text{mol}}{1.000 , \text{kg}}) = (0.263 , \text{m})

Since magnesium chloride dissociates into three ions (Mg2+ and 2Cl-) when dissolved in water, the van't Hoff factor ((i)) is 3.

Now, we can calculate the freezing point depression:

(\Delta T_f = 3 \times 1.86 , \text{°C} \cdot \text{kg/mol} \cdot 0.263 , \text{m})

(\Delta T_f ≈ 1.558 , \text{°C})

Finally, we find the freezing point of the solution by subtracting the freezing point depression from the normal freezing point of water (0°C):

Freezing point of the solution = (0 , \text{°C} - 1.558 , \text{°C})

Freezing point of the solution ≈ -1.558 °C

Therefore, the freezing point of the new solution, assuming complete dissociation of the MgCl2 salt, is approximately -1.558°C.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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