# Answer to Question #201257 in General Chemistry for Koporo

Question #201257

1 what is the percent by mass of oxygen in magnesium oxide , MgO

2 A compound is approximately 86% carbon and 14 % hydrogen by mass. What is the empirical formula for this compound

3 100g of eucalyptol, the chief constituent of eucalyptus oil, consist of 6.484 mol C 11.67 lol H, and 0.6481 mol O

What is the empirical formula of this compound.

4 how many millimetres of 3.0 M H2SO4 are needed to make 450 mL of 0.10 MH2SO4

5 how many moles of HCl are there in 15.0 mo2 of solution with a concentration of 2.50 mol-1

6 If I prepare a solution by adding 75 mL of 1.25 M HCl into a 250 mL volumetric flask, and filling the flask to the mark with water , What will the final concentration of the solution be

7 what is the molarity of glacial acetic acid ( CH3COOH, Mr =60.05 g/mL ) at 25 degrees Celsius given that the density of acetic acid at that temperature is 1.049 g/mL

8 what is the molarity of a KF (aq) solution containing 116.2 g of KF in 3.00 L of solution

1
2021-06-01T07:00:20-0400

Percent of oxygen in MgO

carbon = 86 %

Hydrogen = 14 %

Mass of eucalyptol = 100 g

mol of C =6.484 mol

Mol of H = 11.67 mol

Mol of O = 0.6481 mol

1) Molar mass of Mg = 24 g/mol

Molar mass of O = 16 g/mol

Moles of Mg = 1 and O = 1 in MgO

Therefore mass of Mg = 24 g

Mass of O = 16 g

Total mass = 24 + 16 = 40 g

%O="\\frac{Mass of O}{Total Mass}\u00d7100"

"=\\frac{16}{40}\u00d7100"

%O=40%

2) Let's assume mass of compound = 100 g

Then mass of carbon = 86 g

Mass of Hydrogen = 14 g

Molar mass of carbon = 12 g/mol

Molar mass of hydrogen = 1 g/mol

Moles of Carbon = "\\frac{86}{12} = 7.2 mol"

Moles of Hydrogen = "\\frac{14}{1 }= 14 mol"

Ratio must be whole number so divide smaller number to both value

C =  "\\frac{7.2}{7.2} = 1"

H = "\\frac{14}{7.2 }= 1.9 =2"

Therefore empirical formula for this compound CH2

3)mol of C =6.484 mol

Mol of H = 11.67 mol

Mol of O = 0.6481 mol

Ratio must be whole number so divide smaller number (0.6481) to all the value

C = "\\frac{6.484}{0.6481} = 10"

H = "\\frac{11.67}{0.6481} = 18"

O =  "\\frac{0.6481}{0.6481} = 1"

Therefore empirical formula for this compound C10H18O

(4)The molarity is the concentration term defined as the number of moles of solute dissolved per liter volume of the solution. It is denoted as M. The formula of expression for it is-

"M=\\frac{no. of moles of solute}{Volume of solution in litres}"

The equation in case of dilution is-

C₁V₁ = C₂V₂

Given data is-

C1= 3.0 M

V1= To calculate

C2= 0.10 M

V2= 450 mL= 0.45 L

Putting the given data in the equation of dilution is- Thus, 15 millimetres of 3.0 M H2SO4are needed to make 450 mL of 0.10 M H2SO4.

(5)Given

For HCl solution

volume of solution = 15 ml

Concentration of solution = 2.50 M

Now to calculate the moles of HCl we have to use the formula of molarity

Molarity = "\\frac{moles of solute}{volume of solution (ml)}\u00d71000"

By above formula

Substitute

(2.50) ="\\frac{moles of solute}{ 15 ml}\u00d71000"

moles of solute (HCl) = 0.0375 moles

Moles of HCl are present in the solution = 0.0375 moles

(6)The equation for dilution is : Given : Calculating Molarity of the diluted solution, by using dilution equation : The concentration of the diluted solution is 0.38 M .

(7)Molarity = "\\frac{number moles }{V in ml}\u00d71000"

Number of moles = "\\frac{mass of substance }{molecular mass of substance}"

Mol. Mass of acetic acid is 60.05

Density ="\\frac{mass}{volume}" = 1.049

Or mass = density x volume

Mass = 1.049 x 1litre

=1.049x1000

=1049 is the mass of acetic acid

Mol mass = 60.05

Volume is 1 litre then

Molarity = 1049/60.05 x 1/1L"\\frac{1049}{60.05} \u00d7 1"

Molarity = 17.46M

Thus, the molarity of glacial acetic acid is 17.46 Molar

(8)Given:

The amount of KF in the solution, m = 116.2 g

The volume of the solution, V = 3.00 L

Also, the molar mass of KF (39.0983 + 18.9984), M= 58.0967 g/mol

Molarity (M) of a solution is given as:

"M=\\frac{Moles of solute }{Litres of Solution}"

Firstly, finding the moles of KF as:

"\\frac{116.2g}{58.0967g\/mol}=2.00mol"

Now, the molarity of the KF solution is calculated as:

"M=\\frac{2.00mol}{3.00L}"

"=0.666mol\/L"

Therefore, the molarity of the KF solution is 0.666 M.

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