All categories

3 years ago

What is the freezing point of a 0.82 m aqueous solution of a non-volatile non-electrolyte?

Chemistry

1 answer:

Hoochie [10]3 years ago

6 0

Answer:

-1.5 °C

Explanation:

The freezing-point depression is a colligative property, that is, the property of a solution. The freezing point of a solution is lower than the freezing point of the pure solvent. We can find the freezing-point depression of a solution of <em>a non-volatile non-electrolyte solute</em> using the following expression.

$\Delta T_f = K_f \times b$

where,

$\Delta T_f$ : the freezing-point depression

$K_f$ : cryoscopic constant (For water, Kf = 1.86 °C/m)

$b$ : molality

$\Delta T_f = \frac{1.86\°C}{m} \times 0.82m = 1.5\°C$

The freezing-point of pure water is 0°C. The freezing-point of the solution is:

$0\°C-1.5\°C = -1.5\°C$

You might be interested in

100.00 mL of 0.15 M nitrous acid (HNO2) are titrated with a 0.15 M NaOH solution. (a) Calculate the pH for the initial solution.

wolverine [178]

Answer:

a. pH = 2.04

b. pH = 3.85

c. pH = 8.06

d. pH = 11.56

Explanation:

The nitrous acid is a weak acid (Ka = 5.6x10⁻⁴) that reacts with NaOH as follows:

HNO₂ + NaOH → NaNO₂(aq) + H₂O(l)

a. At the beginning there is just a solution of 0.12M HNO₂. As Ka is:

Ka = [H⁺] [NO₂⁻] / [HNO₂]

Where [H⁺] and [NO₂⁻] ions comes from the same equilibrium ([H⁺] = [NO₂⁻] = X):

5.6x10⁻⁴ = X² / 0.15M

8.4x10⁻⁵ = X²

X = [H⁺] = 9.165x10⁻³M

As pH = -log [H⁺]

b. At this point we have HNO₂ and NaNO₂ (The weak acid and the conjugate base), a buffer. The pH of a buffer is obtained using H-H equation:

pH = pKa + log [NaNO₂] / [HNO₂]

<em>Where pH is the pH of the buffer,</em>

<em>And [NaNO₂] [HNO₂] could be taken as the moles of each compound.</em>

<em />

The initial moles of HNO₂ are:

0.100L * (0.15mol / L) = 0.015moles

The moles of base added are:

0.0800L * (0.15mol / L) = 0.012moles

The moles of base added = Moles of NaNO₂ produced = 0.012moles.

And the moles of HNO₂ that remains are:

0.015moles - 0.012moles = 0.003moles

Replacing in H-H equation:

pH = 3.25 + log [0.012moles] / [0.003moles]

c. At equivalence point all HNO2 reacts producing NaNO₂. The volume added of NaOH must be 100mL. That means the concentration of the NaNO₂ is:

0.15M / 2 = 0.075M

The NaNO₂ is in equilibrium with water as follows:

NaNO₂(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq) + Na⁺

The equilibrium constant, kb, is:

Kb = Kw/Ka = 1x10⁻¹⁴ / 5.6x10⁻⁴ = 1.79x10⁻¹¹ = [OH⁻] [HNO₂] / [NaNO₂]

<em>Where [OH⁻] = [HNO₂] = x</em>

<em />

1.79x10⁻¹¹ = [X] [X] / [0.075M]

1.34x10⁻¹² = X²

X = 1.16x10⁻⁶M = [OH⁻]

pOH = -log [OH-] = 5.94

pH = 14-pOH

d. At this point, 5mL of NaOH are added in excess, the moles are:

5mL = 5x10⁻³L * (0.15mol / L) =7.5x10⁻⁴moles NaOH

In 100mL + 105mL = 205mL = 0.205L. [NaOH] = 7.5x10⁻⁴moles NaOH / 0.205L =

3.66x10⁻³M = [OH⁻]

pOH = 2.44

pH = 14 - pOH

5 0

2 years ago

Express the following numbers as decimals: (a) 1.52 x 10^-2, (b) 7.78 x 10^-8, (c) 1 x 10^-6, (d) 1.6001 x 10^3.

Alchen [17]

Answer :

(a) 0.0152

(b) 0.0000000778

(c) 0.000001

(d) 1600.1

Explanation :

Scientific notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.

For example :

5000 is written as $5.0\times 10^3$

889.9 is written as $8.899\times 10^{-2}$

In this examples, 5000 and 889.9 are written in the standard notation and $5.0\times 10^3$ and $8.899\times 10^{-2}$ are written in the scientific notation.

(a) $1.52\times 10^{-2}$

The standard notation is, 0.0152

(b) $7.78\times 10^{-8}$

The standard notation is, 0.0000000778

(c) $1\times 10^{-6}$

The standard notation is, 0.000001

(d) $1.6001\times 10^{3}$

The standard notation is, 1600.1

5 0

3 years ago

A solid cylinder having a diameter of 1.50 cm and a height of 5.15 cm has a mass of 95.56 g. Show the equations needed to calcul

Georgia [21]

Answer:

you can solve the rest of the equation. I only reduced it to that much to show you how to derive it

4 0

2 years ago

Select the correct Answers.

Mandarinka [93]

Answer:

3. The temperatures of the two substances equalize.

Explanation:

As two objects at different temperatures are placed in contact, heat is transferred from the warmer to the cooler object until the temperature of the two objects be the same.

The amount of heat that is transferred from the warmer object is equal to the amount of heat that is transferred into the cooler object.

This is in agreement with the law of conservation of energy.
<em>So, the right choice is: 3. The temperatures of the two substances equalize. </em>

8 0

3 years ago

Read 2 more answers

21. What is the frequency, given 2-3 x 10¹m? Show all work

Aloiza [94]

Answer:

Frequency is 2Hz

Explanation:

5 0

1 year ago