Pretty please answer this

A solution containing 20.0 g of an unknown liquid and 110.0 g water has a freezing point of .32 °C. Given Kf 1.86°C/m for water,

hjlf

Answer:

A. 256

Explanation:

In a solution where a liquid is the sovent, we'll use the van't Hoff factor, which is the ratio between the number of moles of particles produced in solution and the number of moles of solute dissolved, will be equal to 1.

ΔTemp.f = i * Kf * b

where,

ΔTemp.f = the freezing-point depression;

i = the van't Hoff factor

Kf = the cryoscopic constant of the solvent;

b = the molality of the solution.

So the freezing-point depression by definition is the difference between the the freezing point of the pure solvent and the freesing point of the solution.

Mathematically,

ΔTemp.f = Temp.f° - Temp.f

where,

Temp.f° = the freezing point of the pure solvent.

Temp.f = the freezin point of the solution.

Freezing point of pure water = 0°C

ΔTemp.f = 0 - (-1.32)

= 1.32°C

i = 1,

Kf = 1.86 °Ckg/mol

Solving for the molality, b = ΔTemp.f/( i * Kf)

= 1.32/(1*1.86)

= 0.71 mol/kg

Converting from mol/kg to mol/g,

0.71 mol/kg * 1kg/1000g

= 0.00071 mol/g.

Mass of solvent = 110g

Number of moles = mass * molality

= 0.00071 * 110

= 0.078 mol.

To calculate molar mass,

Molar mass (g/mol) = mass/number of moles

Mass of solute (liquid) = 20g

Molar mass = 20/0.078

= 256.2 g/mol

4 0

3 years ago

A sample of propane, C3Hg, contains 14.4 moles of carbon atoms. How many total moles of atoms does

Leona [35]

Answer:

43.2 mol C

115 mol H

158 mol of atoms

Explanation:

Step 1: Given data

Moles of C₃H₈: 14.4 mol

Step 2: Calculate the number of moles of C

The molar ratio of C₃H₈ to C is 1:3. The moles of C are 3/1 × 14.4 mol = 43.2 mol.

Step 3: Calculate the number of moles of H

The molar ratio of C₃H₈ to H is 1:(. The moles of H are 8/1 × 14.4 mol = 115 mol.

Step 4: Calculate the total number of moles of atoms

n = nC + nH = 43.2 mol + 115 mol = 158 mol

8 0

3 years ago

15. Which compounds are reactants in the process of cellular respiration?

scoundrel [369]

Answer: the correct answer would be the last option, C6H12O6 and O2

Explanation:

5 0

3 years ago

1) Write the ionic compound formulas for the element sodium and all the elements in group 6A. Example: Sodium + Oxygen=Na2O

PIT_PIT [208]

Answer:

See explanation

Explanation:

1) Group 6A elements include; O, S, Se, Te, Po

Na2O, Na2S, Na2Se, Na2Te, Na2Po

2) Group 7A elements include; F, Cl, Br, I, At

AlF3, AlCl3, AlBr3, AlI3, AlAt3

3) Group 5A elements are;

N, P, As, Sb,Bi

Mg3N2, Mg3P2, Mg3As2, Mg3Sb2, Mg3Bi2

3 0

3 years ago

10) How many grams are there in 1.00 x 10^24 molecules of BC13?

lana66690 [7]

194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

Explanation:

In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.

It is known that 1 moles of any element has 6.022×10²³ molecules.

Then 1 molecule will have $\frac{1}{6.022*10^{23} }$ moles.

So $1*10^{24} molecules = \frac{1*10^{24} }{6.022*10^{23} } =1.66 moles$

Thus, 1.66 moles are included in BCl₃.

Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.

As it is known as 1 mole contains molecular mass of the compound.

As the molecular mass of BCl₃ will be

$Molecular mass of BCl_{3}= Mass of Boron + (3*Mass of chlorine)$

Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.

Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.

$grams of BCl_{3}=Molar mass of BCl_{3}*Number of moles$

$grams of BCl_{3}=117.17*1.66=194.5 g$

So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

3 0

4 years ago