Solve and show work. How many grams of potassium permanganate, KMnO4 are need to make 1.25 L of 0.75 M solution?
1 answer:
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Answer:
(a) Moles of ammonium chloride = 0.243 moles
(b)
(c) 60.68 mL
Explanation:
(a) Mass of ammonium chloride = 13.0 g
Molar mass of ammonium chloride = 53.491 g/mol
The formula for the calculation of moles is shown below:
Thus,
<u>Moles of ammonium chloride = 0.243 moles</u>
(b) Moles of ammonium chloride = 0.243 moles
Volume = 295 mL = 0.295 L ( 1 mL = 0.001 L)
(c) Moles of ammonium chloride = 0.0500 moles
Volume = ?
Molarity = 0.824 M
<u>Volume = 0.05 / 0.824 L = 0.06068 L = 60.68 mL</u>