Question

1. Calculate the molarity when 403 g MgSO4 is dissolved to make a 5.25 L solution. Round to two significant digits.

2. Calculate the molarity when 1.24 g AgNO3 is dissolved to make 125 mL solution. Round to two significant digits.

3. How many ml of acetic acid are needed to make 500 mL of 5.0% acetic acid (by volume/volume)

4. What is the morality of a solution of sodium chloride prepared by diluting 750.0 mL of a 6.0 M sodium chloride solution to 2.0 L?

5. How many grams of solute in grams are needed to make 75 mL of 0.160 M C12H22O11.

6. How many grams of naphthalene (C10H8) are needed to make 500 mL of a 1.2 M benzene solution.

7. How many grams of ethanol are in 354 mL (12 oz.) of a 3.2% (mass/mass) beer? Assume 1 mL of beer has a mass of 1 g.

8. The average American has 11 ppm of DDT in their fat tissue (1970 data). If you have 5 Kg of fatty tissue in your body, how many mg of DDT do you contain?

Answer 1

1. Molarity of MgSO₄ = 0.6 M

2. Molarity of AgNO₃ = 0.06 M

3. volume of acetic acid = 250 mL

4. Molarity of NaCl= 2.3 M

5. Mass of C₁₂H₂₂O₁₁ = 4.1 g

6. Mass of C₁₀H₈ (naphthalene) = 77 g

7. mass of ethanol = 11 g

8. Mass of DDT = 0.011 mg

Step-by-step explanation:

Ans 1.

Data given

mass of MgSO₄ = 403

Volume of solution = 5.25 L

Solution:

First we will find mole of solute

no. of moles = mass in grams / Molar mass . . . . . .(1)

Molar mass of MgSO₄ = 24 + 32 + 4(16)

Molar mass of MgSO₄ = 120 g/mol

Put values in equation 1

no. of moles = 403 g / 120 g/mol

no. of moles = 3.36 mol

Now to find molarity

Formula used

Molarity = No. of moles of solute / solution in L . . . . . . (2)

Put Values in above equation

M = 3.36 mol / 5.25 L

M = 0.64

So, round the figure to two significant digits.

Molarity of MgSO₄ = 0.64 M

_____________________

Ans 2.

Data given

mass of AgNO₃ = 1.24 g

Volume of solution = 125 mL

convert mL to L

1000 mL = 1 L

125 mL = 125/1000 = 0.125

Solution:

First we will find mole of solute

no. of moles = mass in grams / Molar mass . . . . . .(1)

Molar mass of AgNO₃ = 108 + 14 + 3(16)

Molar mass of AgNO₃ = 170 g/mol

Put values in equation 1

no. of moles = 1.24 g / 170 g/mol

no. of moles = 0.0073 mol

Now to find molarity

Formula used

Molarity = No. of moles of solute / solution in L . . . . . . (2)

Put Values in above equation

M = 0.0073 mol / 0.125 L

M = 0.06

So, round the figure to two significant digits.

Molarity of AgNO₃ = 0.06 M

______________________

Ans 3.

Data Given:

volume of acetic acid = 500 mL

% solution of acetic acid (M)= 50%

volume of acetic acid needed = ?

Solution:

formula used

percent of solution = volume of solute/ volume of solution x 100

Put Values in above formula

50 % = volume of solute / 500 mL x 100

Rearrange the above equation

volume of solute = 50 x 500 mL /100

volume of solute = 250 mL

So, volume of acetic acid = 250 mL

______________________

Ans 4

Data Given:

Molarity of NaCl (M1) = 6 M

Volume of NaCl (V1) = 750 mL

convert mL to L

1000 ml = 1 L

750 ml = 750/1000 = 0.75 L

Volume of dilute solution (V2)= 2 L

Molarity of dilute solution (M2) = ?

Solution:

Dilution Formula will be used

M1V1 = M2V2 . . . . . . (1)

Put values in equation 1

6 M x 0.75 L = M2 x 2 L

Rearrange the above equation

M2 = 6 M x 0.75 L / 2 L

M2 = 2.25 M

So, round the figure to two significant digits.

Molarity of NaCl= 2.3 M

_________________________

Ans 5.

Data Given:

Molarity of C₁₂H₂₂O₁₁ = 0.16 M

Mass of C₁₂H₂₂O₁₁ (m) = ?

Volume of dilute solution = 75 mL

convert mL to L

1000 ml = 1 L

75 ml = 75/1000 = 0.075 L

Solution:

As we know

Molarity = no.of moles/ liter of solution . . . . . . .(1)

we also know that

no.of moles = mass in grams / molar mass . . . . . (2)

Combine both equation 1 and 2

Molarity = (mass in grams / molar mass) / liter of solution . . . . (3)

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 12(12) +22(1) +11(16)

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 144 +22 + 176

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 342 g/mol

Put values in equation in equation 3

0.16 M = (mass in grams / 342 g/mol) / 0.075 L

Rearrange the above equation

mass in grams = 0.16 mol/L x 0.075 L x 342 g/mol

mass in grams = 4.104 g

So, round the figure to two significant digits.

Mass of C₁₂H₂₂O₁₁ = 4.1 g

____________________

The remaing portion is in attachment