Question

Describe the preparation of:

1)2.20 L of 0.0850 M KMnO4 from the solid reagent. Dissolve _g KMnO4 in enough water to give a final volume of _L

2) 1.60 L of 0.270 M HClO4 , starting with an 8.00 M solution of the reagent. Take _ mL of the 8.00 M reagent and dilute to _ L with water.

3) 300 mL of a solution that is 0.0850 M in I-, starting with MgI2. Dissolve _ g MgI2 in enough water to give a final volume of _ mL.

4)220 mL of 3.70% (w/v) aqueous CuSO4 from a 0.378 M CuSO4 solution. Take _ mL of the 0.378 M CuSO4 solution and dilute to _ mL with water.

5) 2.20 L of 0.286 M NaOH from the concentrated commercial reagent [50% NaOH (w/w), sp gr 1.525].

Take _ mL of the concentrated reagent and dilute to _ L with water.

Answer 1

We use following formula for making given solution.

Mol = Mass in g / molar mass,

molarity = Moles of solute / volume of solution in L.

Mass percent or (w/w) = (Mass of solute / Mass of solution ) x 100 % , here mass of both solute and solution in same unit.

w/V   here mass of solute is w and V is for volume of solution.

Mass of solute / volume of solution x 100

Dilution formula

M1V1= M2 V2 , M1 is initial molarity , V1 is initial volume , M2 and V2 are the final molarity and volume respectively.

1).

Given:

            Volume of KMnO4 = 2.20 L. [KMnO4]= 0.0850 M

0.0850 M = mol of KMnO4 / 2.20 L

Mass of KMnO4 = (0.0850 M x 2.20 L ) x 158.034 g / mol = 29.55 g

2).

Let M1 = 8.0 M , V1 = unknown

M2 = 0.270 M , V2 = 1.60 L

V1= 0.270 M x 1.60 L / 8.0 M = 0.054 L

So we need 0.054 L = 54.0 mL of 8.0 M and diluted to 1.60 L

3)

Mole ratio of I - : MgI2   is 2:1

We have to moles of I-

Moles of I- = 0.300 L x 0.0850 M = 0.0255 mol I-

Moles of MgI2 = 0.0255 mol I- x 1 mol MgI2 / 2 mol I- = 0.01275 mol MgI2

Mass of MgI2 = 0.01275 mol x molar mass of MgI2 = 0.01275 mol MgI2 x 278.114 g / mol= 3.54 g

Final volume should be 300 mL

4)

Here we use

M1= 0.378 M   , V1 = unknown

M2 = 3.70 % ( w /v) , V2 = 220 mL

Here we convert Molar solution to w/v % solution.

M1 = (0.378 mol/L ) x (1 L / 10³mL ) x Molar mass x 100 %

M1 = (0.378 mol/L ) x (1 L / 10³mL ) x 159.609 g /mol x 100 % = 6.033 w/v %

V1 = 3.70 * 220 mL / 6.033 = 134.92 mL

So the volume of copper sulfate = 134.061 mL and diluted to 220 mL