CHEMISTRY HIGH SCHOOL

PLEASE HELP!!A solution is made with 23.50 g of calcium chloride, CaCI2 and 250.0 mL of water. The water is 25.0°C. What is the freezing point of this solution?

Assume the CaCI2 dissolves completely in the water.. The density of water at 25.0°C is 0.997 g/L, and the KF of H20 is 1.858°C kg/mol.

Answers

Answer 1
Answer:

Answer:

-4741 °C

Something is strange, because this is a weird number.

Explanation:

ΔT = Kf . m. i

That's the colligative property of freezing point depression.

Kf = Cyroscopic constant

m = molality (moles of solute in 1kg of solvent)

i = Van't Hoff factor (numbers of ions dissolved)

We assume 100% dissociation:

CaCl₂ → Ca²⁺ + 2Cl⁻   i = 3

ΔT = Freezing point of pure solvent - Freezing point of solution

Let's determine molality

Solute = CaCl₂

Moles of solute = 23.5 g . 1 mol/ 110.98 g = 0.212 moles

We determine the mass of solvent by density. Density's data is in g/L. We need to convert the volume from mL to L

250 mL . 1L / 1000 mL = 0.250 L

0.997 g/L = mass of water / volume of water → 0.997 g/L . volume of water = mass of water

0.997 g/L . 0.250L = 0.249 g

Now, we convert the mass of water from g to kg

0.249 g . 1 kg / 1000 g = 2.49×10⁻⁴ kg

Molality = mol/kg → 0.212 mol / 2.49×10⁻⁴ kg = 850.5 m

We replace data:

0°C - Freezing point of solution = 1.858 °C . kg /mol . 850.5 mol/kg . 3

Freezing point of solution = -4741 °C

Answer 2
Answer:

Answer:

The freezing point of the solution is -4.74 °C

Explanation:

Step 1: data given

Mass of calcium chloride CaCl2 = 23.50 grams

Step 2: Calculate moles CaCl2

Moles CaCl2 = mass CaCl2 / molar mass CaCl2

Moles CaCl2 = 23.50 grams / 110.98 g/mol

Moles CaCl2 = 0.212 moles

Step 3: Calculate mass H2O

Density = mass / volume

Mass = density * volume

Mass H2O = 997 g/L * 0.250 L

MAss H2O = 249.25 grams

Step 3: calculate molality of the solution

Molality = moles CaCl2 / mass H2O

Molality = 0.212 moles / 0.24925 kg

Molality = 0.851 molal

ΔT = i*Kf*m  

⇒with ΔT = the freezing point depression = TO BE DETERMNED

⇒ with i = the van't Hoff factor = 3

⇒with Kf = the freezing point depression constant of water = 1.858 °C /m

⇒with m = the molality = moles CaCl2 / mass water = 0.851 molal

ΔT = 3 * 1.858 * 0.851

ΔT =  4.74 °C

 

Step 4: Calculate the freezing point of the solution

ΔT = T (pure solvent) − T (solution)

ΔT  = 0°C - 4.74 °C

The freezing point of the solution is -4.74 °C

NOTE: the density of water = 0.997 kg/L or 997 g/L

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C is the correct answer

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Answers

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What are the thin fibers of the human nervous system that send messages between different parts of the body?A.
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Answers

Answer:

The correct answer is option A, that is, nerves.

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The nerves are also known as neurons that make up the nervous system. The nerves are considered as the unique cells, which conduct messages from one section of the body to another, as minute electrical signals. These messages are also called nerve impulses. There are three distinct kinds of neurons known, and each one exhibit a somewhat different function. These are relay neurons, sensory neurons, and motor neurons.
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Answers

Answer:

The correct option is: Potassium

Explanation:

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Answers

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Answers

Answer:

Option B. Add 1.46 g of NaCl to 250 mL of H₂O

Explanation:

Analyse the data given: 250 mL and 0.10 M

The molar concentration and the volume of solution can indicate the moles of solute that were used.

Let's convert the volume from mL to L → 250 mL . 1 L /1000mL =0.250L

M . volume = moles → 0.10 mol/L . 0.250L = 0.025 moles

If we convert the moles to mass, we'll know the mass of solute:

0.025 mol . 58.4 g/mol = 1.46 g

Answer:

We have to solve 1.46 grams of NaCl in 250 mL H2O (option B)

Explanation:

Step 1: Data given

Volume = 250 mL = 0.250 L

Molarity NaCl solution = 0.10 M

Molar mass of NaCl = 58.4 g/mol

Step 2: Calculate moles NaCl

Moles NaCl = molarity * volume

Moles NaCl = 0.10 M * 0.250 L

Moles NaCl = 0.025 moles

Step 3: Calculate mass NaCl

Mass NaCl = moles NaCl * molar mass NaCl

Mass NaCl = 0.025 moles * 58.4 g/mol

Mass NaCl = 1.46 grams

We have to solve 1.46 grams of NaCl in 250 mL H2O (option B)
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