Question

8) What is the molarity (M) concentration of a solution when 18.37 grams of potasslum nitrite (KNO,) is dissolved in water to make 120 mL of solution ? GENERAL CHEMISTRY LABORATORY II 9) What will the final molarity concentration of the solution in question 8 if the volume is doubled?

Answer 1

8)

Mass of potassium nitrite = 18.37 g

Molar mass of potassium nitrite = 85.104 g/ mol

Moles = mass/molar mass

Moles = 18.37 g / 85.104 g /mol

Moles = 0.216 moles

Volume in L = 0.120 L

Molarity = number of moles / Volume in L

Molarity = 0.216 moles / 0.120 L

Molarity = 1.8 M

----------------------------------------------------------------------------------------

9)

Volume is doubled = 0.240 L

Molarity = number of moles / Volume in L

Molarity = 0.216 moles / 0.240 L

Molarity = 0.9 M

The molarity is halved when volume is doubled.

----------------------------------------------------------------------------------

10)

Mass of Iron sulfate = 6.44 g

Molar mass of Iron sulfate = 151.908 g/mol

Moles of iron sulfate = mass / molar mass

Moles = 6.4 g / 151.908 g/mol

Moles = 0.0424 moles

According to the equation

10 moles of iron sulfate react with = 2 moles of KMnO4

1 mole of iron sulfate react with = 2/10 moles of KMnO4

0.0424 moles of iron sulfate react with = 0.0424*0.2 moles of KMnO4

Moles of KMnO4 = 0.0085 moles

Molarity = 0.153 moles/L

Volume of KMnO4 = moles /molarity

Volume = 0.0085 moles / 0.153 moles /L

Volume = 0.05556 L

Volume = 55.56 mL

Therefore, the volume is 55.56 mL.