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?

Question

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

Caleb Adams · Accepted Answer

Here's what I got.

Extensive Response Magnesium chloride, #"MgCl"_2#, is a soluble ionic compound that dissociates in aqueous solution to form magnesium cations, #"Mg"^(2+)#, and chloride anions, #"Cl"^(-)# #"MgCl"_text(2(aq]) -> "Mg"_text((aq])^(2+) + 2"Cl"_text((aq])^(-)# 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. As you know, the freezing point of a solution depends on how many particles of solute you have present, not on the nature of the solute #-># colligative property. The freezing-point depression of a solution can be expressed mathematically using the following equation: #color(blue)(DeltaT_f = i * K_f * b)" "#, where #DeltaT_f# - the freezing-point depression; #i# - the van't Hoff factor #K_f# - the cryoscopic constant of the solvent; #b# - the molality of the solution. The cryoscopic constant of water is equal to #1.86 ""^@"C kg mol"^(-1)# * 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. Since you know that every mole of magnesium chloride produces three moles of ions in solution, you can say that the van't Hoff factor will be equal to #3#. In order to find the molality of the solution, you need to know how many moles of solute you have in that #"25.0-g"# sample. Use the compound's molar mass to accomplish that. #25.0 color(red)(cancel(color(black)("g"))) * "1 mole MgCl"_2/(95.21color(red)(cancel(color(black)("g")))) = "0.2626 moles MgCl"_2# Molality is now defined as the number of solute moles in one kilogram of solvent. #color(blue)(b = n_"solute"/m_"solvent")# In your situation, you'll have #b = "0.2626 moles"/(1000 * 10^(-3)"kg") = "0.2626 molal"# Accordingly, the freezing-point depression is expected to be #DeltaT_f = 3 * 1.86^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.2626color(red)(cancel(color(black)("mol")))color(red)(cancel(color(black)("kg"^(-1)))) = 1.465^@"C"# The definition of the freezing-point depression is #color(blue)(DeltaT_f = T_f^@ - T_"f sol")" "#, where #T_f^@# - the freezing point of the pure solvent #T_"f sol"# - the freezing point of the solution This indicates that you've #T_"f sol" = T_f^@ - DeltaT_f# #T_"f sol" = 0^@"C" - 1.465^@"C" = -1.465^@"C"# 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. #T_"f sol" = color(green)(-1.5^@"C")# This brings us to the boiling point of this solution: the boiling-point elevation equation looks like this. #color(blue)(DeltaT_b = i * K_b * b)" "#, where #DeltaT_b# - the boiling-point elevation; #i# - the van't Hoff factor #K_b# - the ebullioscopic constant of the solvent; #b# - the molality of the solution. The ebullioscopic constant for water is equal to #0.512^@"C kg mol"^(-1)# * Tcriosebu.pdf https://tutor.hix.ai Enter your values to obtain #DeltaT_b = 3 * 0.512^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.2626 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1)))) = 0.403^@"C"# The definition of the boiling-point elevation is #color(blue)(DeltaT_b = T_"b sol" - T_b^@)" "#, where #T_"b sol"# - the boiling point of the solution #T_b^@# - the boiling point of the pure solvent As for you, you've #T_"b sol" = DeltaT_b + T_b^@# #T_"b sol" = 0.403^@"C" + 100^@"C" = 100.403^@"C"# This response will remain as #T_"b 'sol" = color(green)(100.4^@"C")#

Naomi Aronson · Answer

undefined 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.