Question

QUESTION 1

What volume of 0.10 mol/L sulfuric acid is needed to create 50 mL of a 0.080 mol/L solution?

		50 mL
		40 mL
		1 L
		10 mL

20 points   

QUESTION 2

Solution 1 contains Malonic acid and 0.020 mol/L Manganese (II) sulfate monohydrate. What mass of Manganese (II) sulfate monohydrate (MW = 169 g/mol) should you weigh to prepare a final volume of 25.0 mL of solution 1?

		0.084 g of MnSO4. H2O
		140 g of MnSO4. H2O
		0.0030 g of MnSO4. H2O
		34 g of MnSO4. H2O

20 points   

QUESTION 3

Solution 1 contains 0.15 mol/L Malonic acid and Manganese (II) sulfate monohydrate. What mass of malonic acid (C₃H₄O₄) should you weigh to prepare a final volume of 25.0 mL of solution 1?

		0.0039 mg of Malonic acid
		0.10 g of Malonic acid
		0.39 g of Malonic acid
		10 g of Malonic acid

20 points   

QUESTION 4

Solution 2 contains 0.20 M potassium iodate dissolved in 0.080 M H₂SO₄? What mass of potassium iodate (KIO₃) should you weigh to prepare a final volume of 25.0 mL of solution 2?

		1.1 g of KIO3
		0.16 g of KIO3
		0.08 g of KIO3
		0.21 g of KIO3

20 points   

QUESTION 5

Solution 2 contains 0.20 M potassium iodate dissolved in 0.080 M H₂SO₄? What volume of 0.080 M H₂SO₄ should you add to potassium iodate in order to prepare a final volume of 25.0 mL of solution 2?

		80.0 mL
		25.0 mL
		20.0 mL
		40.0 mL

Answer 1

1)

use dilution formula

M1*V1 = M2*V2

1---> is for stock solution

2---> is for diluted solution

Given:

M1 = 0.1 M

M2 = 0.08 M

V2 = 50 mL

use:

M1*V1 = M2*V2

V1 = (M2 * V2) / M1

V1 = (0.08*50)/0.1

V1 = 40 mL

Answer: 40 mL

2)

volume , V = 25 mL

= 2.5*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 2*10^-2*2.5*10^-2

= 5*10^-4 mol

use:

mass,

m = number of mol * molar mass

= 5*10^-4*1.69*10^2

= 8.45*10^-2 g

Answer: 0.084 g

Only 1 question at a time please