1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
denis23 [38]
3 years ago
14

A certain liquid has a normal freezing point of and a freezing point depression constant . Calculate the freezing point of a sol

ution made of of iron(III) chloride () dissolved in of . Round your answer to significant digits.
Chemistry
1 answer:
katrin2010 [14]3 years ago
6 0

The question is incomplete, the complete question is:

A certain liquid X has a normal freezing point of 0.80^oC and a freezing point depression constant K_f=7.82^oC.kg/mol . Calculate the freezing point of a solution made of 81.1 g of iron(III) chloride () dissolved in 850. g of X. Round your answer to significant digits.

<u>Answer:</u> The freezing point of the solution is -17.6^oC

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m

OR

\text{Freezing point of pure solvent}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}} ......(1)

where,

Freezing point of pure solvent = 0.80^oC

Freezing point of solution = ?^oC

i = Vant Hoff factor = 4 (for iron (III) chloride as 4 ions are produced in the reaction)

K_f = freezing point depression constant = 7.82^oC/m

m_{solute} = Given mass of solute (iron (III) chloride) = 81.1 g

M_{solute} = Molar mass of solute (iron (III) chloride) = 162.2 g/mol

w_{solvent} = Mass of solvent (X) = 850. g

Putting values in equation 1, we get:

0.8-(\text{Freezing point of solution})=4\times 7.82\times \frac{81.1\times 1000}{162.2\times 850}\\\\\text{Freezing point of solution}=[0.8-18.4]^oC\\\\\text{Freezing point of solution}=-17.6^oC

Hence, the freezing point of the solution is -17.6^oC

You might be interested in
Latent heat is used to: 1. form chemical bonds. 2. change molecular structure. 3. change states of matter. 4. change the tempera
likoan [24]

Answer : The correct option is, (3) change states of matter.

Explanation :

Latent heat : It is defined as the heat required to convert the solid into liquid or vapor and a liquid into a vapor without changing the temperature.

There are two types of latent heat.

(1) Latent heat of fusion

(2) Latent heat of vaporization

Latent heat of fusion : It is defined as the amount of heat energy released or absorbed when the solid converted to liquid at atmospheric pressure at its melting point.

Latent heat of vaporization : It is defined as the amount of heat energy released or absorbed when the liquid converted to vapor at atmospheric pressure at its boiling point.

Hence, latent heat is used to change states of matter.

7 0
3 years ago
What mass of NaC6H5COO should be added to 1.5 L of 0.40 M C6H5COOH solution at 25 °C to produce a solution with a pH of 3.87 giv
statuscvo [17]

Answer:

41 g

Explanation:

We have a buffer formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ coming from NaC₆H₅COO). We can find the concentration of C₆H₅COO⁻ (and therefore of NaC₆H₅COO) using the Henderson-Hasselbach equation.

pH = pKa + log [C₆H₅COO⁻]/[C₆H₅COOH]

pH - pKa = log [C₆H₅COO⁻] - log [C₆H₅COOH]

log [C₆H₅COO⁻] = pH - pKa + log [C₆H₅COOH]

log [C₆H₅COO⁻] = 3.87 - (-log 6.5 × 10⁻⁵) + log 0.40

[C₆H₅COO⁻] = [NaC₆H₅COO] = 0.19 M

We can find the mass of NaC₆H₅COO using the following expression.

M = mass NaC₆H₅COO / molar mass NaC₆H₅COO × liters of solution

mass NaC₆H₅COO = M × molar mass NaC₆H₅COO × liters of solution

mass NaC₆H₅COO = 0.19 mol/L × 144.1032 g/mol × 1.5 L

mass NaC₆H₅COO = 41 g

7 0
3 years ago
Chemical formula for Potassium Hydrogen Nitrate
My name is Ann [436]

Answer:

KNO3

Explanation:

8 0
3 years ago
PLEASE HURRY UP!! I DON'T HAVE ALL DAY!
neonofarm [45]

B: The total thermal energy is greater in a large body of water than one much smaller

Explanation:

A large lake filled filled with cool water will have more thermal energy than smaller pond filled with warmer water because the total thermal energy is greater in a large body of water than one that is much smaller.

Thermal energy is a form of kinetic energy usually due to transfer of heat energy.

Amount of heat energy is dependent on the differences in temperature, mass and specific heat capacity of a body.

Both lake water will have the same specific heat capacity. Since larger body of water has more mass, it will possess more thermal energy.

learn more:

Specific heat capacity brainly.com/question/7210400

Thermal energy brainly.com/question/914750

#learnwithBrainly

7 0
3 years ago
Identify the compound. Which of these is a compound?
Bond [772]
The answer has to be either CO2 or mg. But i am not sure which one 
7 0
3 years ago
Read 2 more answers
Other questions:
  • If you want to raise the temperature of 200 grams of water at 20 Celsius to 30 Celsius how much water and at what temperature wo
    14·1 answer
  • What are the three steps scientists take to evaluate a scientific explanation
    14·1 answer
  • Balance the equation: ___HgO → ___O2 + ___Hg
    5·2 answers
  • What is the chemical characteristics of the compound barium oxide?
    7·1 answer
  • You need 2.5 moles of aluminum for an experiment. How many atoms of aluminum is this?​
    14·1 answer
  • I need help plzzz help me!!!
    12·1 answer
  • Susan and her friend found a piece of metal which they want to
    7·1 answer
  • What do solar panels convert radiant energy into?
    7·2 answers
  • What happen when molten zinc bromide is electrolysised
    14·1 answer
  • At 1 atmosphere and 20 degrees Celsius, all samples of H20(l) must have the same
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!