Question

Determine how many grams of each of the following solutes would be needed to make 3.40*10² mL of a 0.100 M solution

A) Cesium iodide (CsI)

B)Sulfuric acid

C)sodium carbonate

D)Potassium dichromate

E)potassium permanganate

Answer 1

Here we have to use the molarity equation to calculate the amount of given molecules.

Here w= ?

MM= molar mass in gm/mol

v= volume=3.42 x 10² mL=340 mL

molarity = 0.100 M

A) MM of Cesium iodide (CsI)= 259.81 gm/mol

v= 340 mL

molarity= 0.1 M

now apply

Hence 0.88 gm CsI is required.

B)

MM of Sulfuric acid = 98.07 gm/mol

v= 340 mL

molarity= 0.1 M

now apply

Hence 0.33 gm CsI is required.

C)

MM of Sodium carbonate = 106 gm/mol

v= 340 mL

molarity= 0.1 M

now apply

Hence 0.36 gm CsI is required.

D)

MM of Potassium dichromate = 294.18 gm/mol

v= 340 mL

molarity= 0.1 M

now apply

Hence 1 gm CsI is required.

E)

MM of Potassium permanganate= 158 gm/mol

v= 340 mL

molarity= 0.1 M

now apply

Hence 0.53 gm CsI is required.

Answer.