Chemistry College

Determine whether each of the examples represents a colligative property or non colligative property
1. freezing point depression
2. density
3. boiling point elevation
4. vapor pressure lowering
5. colour

Answers

Answer 1

After Determining each examples, we found that:

Freezing point depression - Colligative propertyDensity - Non colligative propertyBoiling point elevation - Colligative propertyVapor pressure lowering - Colligative propertyColor - Non colligative property

Colligative properties are properties of a solution that depend on the number of solute particles present in the solvent, but not on the chemical identity of the solute. These properties include freezing point depression, boiling point elevation, and vapor pressure lowering.

Non-colligative properties, on the other hand, are properties of a solution that depend on the chemical identity of the solute. They are not dependent on the number of solute particles in the solvent. An example of a non-colligative property is density.

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Related Questions

How many grams of potassium
carbonate are needed to make
400.0 mL of 2.5 M solution?

K₂CO3; 138.21 g/mol



A. 21 g K₂CO3
C. 140 g K₂CO3
B. 140,000 g K₂CO3
D. 86 g K₂CO3

Answers

The grams of potassium carbonate neede to make 400ml of 2.5M solution is 186g.

How to calculate grams?

A answer's molarity tells you approximately the variety of moles of solute this is found in 1 L of answer. You can say:

Molarity(M)=moles of solute1 litre of answer Molarity(M)=moles of solute/litre of answer

We recognise that with a purpose to calculate a answer's molarity, you must recognise the variety of moles of solute this is found in 1 L i.e. a thousand mL of answer.

We can use the molar mass of a compound to decide the variety of moles gift withinside the pattern so variety of moles of a compound may be calculated via way of means of making use of the subsequent system:

Number of moles=Mass(g)Molar mass(gmol−1)Number of moles=Mass(g)/Molar mass(gmol−1)

In the given question, we already recognise the molarity in addition to quantity of the answer that are as follows:

Molarity=2.5Volume=400mL=0.4LMolarity=2.5Volume=400mL=0.4L

Substituting those given values withinside the system of molarity, we can calculate the variety of moles as depicted below:

1 =moles of solute0.4 variety of moles=1 moles of K2CO32=moles of solute0.2 variety of moles=1 moles of K2CO3

Molar mass of K2CO3K2CO3 may be calculated via way of means of including the atomic mass of every of the atom gift withinside the molecule as proven below:

Molecular mass of K2CO3=(2×K)+(1×C)+(3×O)=(2×39)+(1×12)+(3×16)=138gmol−1Molecular mass of K2CO3=(2×K)+(1×C)+(3×O)=(2×39)+(1×12)+(3×16)=138gmol−1

Now substituting the values withinside the variety of moles system, we get the mass of K2CO3K2CO3 as proven below:

1=Mass(g)138Mass(g)=69g0.five=Mass(g)138Mass(g)=138g

Hence, 138 grams of potassium carbonate K2CO3K2CO3 are had to make four hundred mL of a 2.five M answer.

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Answer: C. 140 g K2CO3

Explanation:

beryllium oxide contains 0.5633 g of beryllium for every 1.000 g of oxygen. in 1842 russian chemist avdeev proposed the formula for beryllium oxide. before that time, it was thought that the formula was . assume that oxygen has an atomic mass of 16.00.

Answers

According to the query, they will create an ionic bond by sharing electrons to fill out their respective eight valence electrons. BeO

Here are 2.228 moles of beryllium oxide, or BeO. BeO is stated as the chemical formula for beryllium oxide. The amount of beryllium oxide that was produced weighs 555.7 grams. To determine the number of moles, The moles are calculated by dividing the substance's mass by its molar mass. Molar mass = Mass formed 55.7 grams of BeO were produced in total. BeO has a molar mass of 25g/mol. Assigning values, BeO moles = 55.7/25 BeO is made up of 2.228 moles. Therefore, 2.228 moles of beryllium oxide were produced. The ability of an element to combine with other atoms to form chemical compounds or molecules is measured by the element's valence or valency.

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Liquid ammonia has a vapor pressure of 109 mm Hg at -66C and a heat of vaporization of 2.46 x 10^4 J/mol.

Estimate the normal boiling point of ammonia.

Answers

It is stated in this query that p above p is the rule when using the Clausius-Claparon equation.

How hot does it get when it boils?

According to what I'm going to explain, p not will be our boiling point, and that value will be equal to delta minus. The boiling point is determined by the formula: H, vapor pressure over r 1 over t minus 1 over t not and again t not. Thus, a pressure line can be seen here. According to what we've been told, it is 1 o 9 miles per hour. 680 would enter there, followed by delta h, at a distance of 6080000000 meters from Mercury. The vapourization factor is 2.46. I'm not going to include the units in this section because ideal gas has a constant 8.314 joules per mole.

It is - 66 degrees Celsius when mole kelvin is subtracted from the initial temperature. Since our boiling point would be 2 o 7 kelvin minus 1 over t, that would convert to that value. This will be plugged into my equation using the solver here, which yields 237 kelvin when accounting for t. The boiling point at 680 millimeters of mercury would be minus 273, which gives me a result of minus 36 degrees celsius, and we then translated that to degrees celsius.

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