What is an examole of mass

Camphor, a white solid with a pleasant odor, is extracted from the roots, branches, and trunk of the camphor tree. Assume you di

katrin [286]

Answer: The molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

Explanation:

Calculating the molarity of solution:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of camphor = 70.0 g

Molar mass of camphor = 152.2 g/mol

Volume of solution = 575 mL

Putting values in above equation, we get:

$\text{Molarity of camphor}=\frac{70\times 1000}{152.2\times 575}\\\\\text{Molarity of camphor}=0.799M$

Calculating the molarity of solution:

To calculate the mass of ethanol, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of ethanol = 0.785 g/mL

Volume of ethanol = 575 mL

Putting values in above equation, we get:

$0.785g/mL=\frac{\text{Mass of ethanol}}{575mL}\\\\\text{Mass of ethanol}=(0.785g/mL\times 575mL)=451.38g$

To calculate the molality of solution, we use the equation:

$\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

$m_{solute}$ = Given mass of solute (camphor) = 70 g

$M_{solute}$ = Molar mass of solute (camphor) = 152.2 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 451.38 g

Putting values in above equation, we get:

$\text{Molality of camphor}=\frac{70\times 1000}{152.2\times 451.38}\\\\\text{Molality of camphor}=1.02m$

Calculating the mole fraction of camphor:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For camphor:

Given mass of camphor = 70 g

Molar mass of camphor = 152.2 g/mol

Putting values in equation 1, we get:

$\text{Moles of camphor}=\frac{70g}{152.2g/mol}=0.459mol$

For ethanol:

Given mass of ethanol = 451.38 g

Molar mass of ethanol = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of ethanol}=\frac{451.38g}{46g/mol}=9.813mol$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

Moles of camphor = 0.459 moles

Total moles = [0.459 + 9.813] = 10.272 moles

Putting values in above equation, we get:

$\chi_{(camphor)}=\frac{0.459}{10.272}=0.045$ \

Calculating the mass percent of camphor:

To calculate the mass percentage of camphor in solution, we use the equation:

$\text{Mass percent of camphor}=\frac{\text{Mass of camphor}}{\text{Mass of solution}}\times 100$

Mass of camphor = 70 g

Mass of solution = [70 + 451.38] = 521.38 g

Putting values in above equation, we get:

$\text{Mass percent of camphor}=\frac{70g}{521.38g}\times 100=13.43\%$

Hence, the molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

3 0

3 years ago

4. The reaction of silver nitrate and potassium bromide yields silver bromide and potassium nitrate. If

Hatshy [7]

Answer:

1.) AgNO₃

2.) 0.563 moles AgBr

Explanation:

The limiting reagent is the reagent that is used up completely during a reaction. It can be identified by calculating which reactant produces the smallest amount of product. This can be done by determining the number of moles of each reagent (via molarity conversion). and then converting it to moles of the product (via mole-to-mole ratio).

AgNO₃ (aq) + KBr (aq) ---> AgBr (s) + KNO₃ (aq)

Molarity (M) = moles / liters

100 mL = 1 L

AgNO₃

45.0 mL / 100 = 45.0 L

1.25 M = ? moles / 0.450 L

? moles = 0.563 moles

KBr

75.0 mL / 100 = 0.750 L

0.800 M = ? moles / 0.750 L

? moles = 0.600 moles

In this case, there is no need to use the mole-to-mole ratio because all of the coefficients are one in the reaction (the amount of the limiting reagent used is the same amount of product produced). Since AgNO₃ produces the smaller amount of product, it is the limiting reagent.

4 0

2 years ago

An atom that has 13 protons and 15 neutrons is isotope of the element A.) nickel B.) silicon C.) aluminum D.) phosphorus. Why?

joja [24]

An atom that has 13 protons and 15 neutrons is isotope of Aluminium (answer C)

Explanation

Isotope is a form of the same element with the equal number of protons but difference number of neutrons in their nuclei.

In other words isotope has the same atomic number but different mass number.

Atomic number of a element is determined by number of protons of an element.

from the periodic table Aluminum in atomic number 13 therefore it has 13 protons therefore an atom that has 13 protons and 15 neutrons is a isotope of Aluminium.

6 0

3 years ago

Read 2 more answers

I have no idea on how to do this here is the question (For the following pairs of atoms, show how they transfer valence electron

Maru [420]

Answer:

uhh

Explanation:

i hope you find this answer

3 0

3 years ago

Which is an example of diffraction?

jeyben [28]

I would say d but I’m not sure

8 0

3 years ago