All categories

3 years ago

What is the freezing point of a 0.743m aqueous solution if KCI? (Report amount to three decimal points)

1 answer:

5 0

This is a question about the colligative property known as freezing point depression. Freezing point depression (the amount the normal freezing point of the solvent is decreased) can be calculated with this equation:

ΔT = i Kf m

Where i (the van't Hoff factor) is the degree of dissociation of the solute, Kf is the freezing point depression constant, and m is the molality of the solution.

Here i = 2 (KCl dissociates into 2 ions, K+ and Cl-), Kf = 1.86 C/m (for water), and m = 0.743m).

ΔT = 2 x 1.86 C/m x 0.743m = 2.764C

That means the freezing point of the solution is 2.764C less than the pure solvent (water), making it 0C - 2.764C = -2.764C.

You might be interested in

Which of these pair of elements has the GREATEST electronegativity:

Georgia [21]

I think it might be Selenium.

3 0

2 years ago

Read 2 more answers

Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO 3 ( s ) . The equation for the

Pavlova-9 [17]

Answer:

Mass of $O_2$ produced = 32 g

Explanation:

Calculation of the moles of $KClO_3$ as:-

Mass = 82.4 g

Molar mass of $KClO_3$ = 122.55 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{82.4\ g}{122.55\ g/mol}$

$Moles= 0.67237\ mol$

From the reaction shown below:-

$2KClO_3\rightarrow 2KCl+3O_2$

2 moles of potassium chlorate on reaction forms 3 moles of oxygen gas

So,

0.67237 moles of potassium chlorate on reaction forms $\frac{3}{2}\times 0.67237$ moles of oxygen gas

Moles of oxygen gas = 1 mole

Molar mass of oxygen gas = 32 g/mol

Mass of $O_2$ produced = 32 g

6 0

3 years ago

What is the volume of a cone with a height of 27 cm

andreev551 [17]

Answer:

the answer is 4778.36

Explanation:

do pi time the radius squared times the height divided by 3

5 0

2 years ago

Calculate the mass of ammonium sulfide (nh4)2s in 3.00 l of a 0.0200 m solution. g (nh4)2s

mezya [45]

Taking into account the definition of molarity, 4.08 grams of ammonium sulfide are present in 3.00 L of a 0.0200 M solution.

<h3>Definition of molarity</h3>

Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:

$Molarity=\frac{number of moles}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$ .

<h3>Mass of ammonium sulfide</h3>

In this case, you know:

Molarity: 0.0200 M
number of moles: ?
volume: 3 L

Replacing in the definition of molarity:

$0.0200 \frac{moles}{liter}=\frac{number of moles}{3 L}$

Solving:

0.0200 $\frac{moles}{liter}$ × 3 L= number of moles

0.06 moles= number of moles

On the other side, the molar mass is the amount of mass that a substance contains in one mole. In this case, the molar mass of ammonium sulfide is 68 g/mole.

Then, the amount of mass contained in 0.06 moles is calculated as:

0.06 moles× $\frac{68 grams}{1 mole}$ = 4.08 grams