If 31.6 g of KMnO4 is dissolved in enough water to give 160 mL of solution, what is the molarity?

AND

what mass of oxalic acid, H2C204, is required to make 300 mL of a. 74 M solution?

A. 1.25M

B. 19.98g

Explanation:

A. Data obtained from the question include the following:

Mass of KMnO4 = 31.6 g

Volume = 160 mL

Molarity = ... ?

We'll begin by calculating the number of mole KMnO4 in the solution. This is can be obtained as follow:

Mass of KMnO4 = 31.6 g

Molar mass of KMnO4 = 39 + 55 + (16x4) = 158g/mol

Number of mole of KMnO4 = ... ?

Mole = mass / Molar mass

Number of mole of KMnO4 = 31.6/158 = 0.2 mole

Now, we can obtain the molarity of the solution as follow:

Volume = 160 mL = 160/1000 = 0.16L

Mole of KMnO4 = 0.2 mole

Molarity = mole / Volume

Molarity = 0.2/0.16 = 1.25M

B. Data obtained from the question include the following:

Volume = 300mL

Molarity = 0.74 M

Mass of H2C2O4 = ... ?

First, we shall determine the number of mole H2C2O4. This is illustrated below:

Volume = 300mL = 300/1000 = 0.3L

Molarity = 0.74 M

Mole of H2C2O4 = ?

Mole = Molarity x Volume

Mole of H2C2O4 = 0.74 x 0.3

Mole of H2C2O4 = 0.222 mole

Now, we can easily find the mass of H2C2O4 by converting 0.222 mole to grams as shown below:

Number of mole of H2C2O4 = 0.222 mole

Molar mass of H2C2O4 = (2x1) + (12x2) + (16x4) = 2 + 24 + 64 = 90g/mol

Mass of H2C2O4 = ... ?

Mass = mole x molar mass

Mass of H2C2O4 = 0.222 x 90

Mass of H2C2O4 = 19.98g