A stock solution containing Mn2+ ions was prepared by dissolving 1.748 g pure manganese metal in nitric acid and diluting to a final volume of 1.000 L. The following solutions were then prepared by dilution:
For solution A, 50.00 mL of stock solution was diluted to 1000.0 mL.
For solution B, 10.00 mL of solution A was diluted to 250.0 mL.
For solution C, 10.00 mL of solution B was diluted to 500.0 mL.
Calculate the concentrations of the stock solution and solutions A, B, and C.
2007-09-22 04:49:51 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
firstly write out the equation of the reaction between manganese metal and nitric acid.
Mn(s) + 2HNO3(aq) ---> Mn2+(aq) + 2NO3- (aq) + H2 (g)
thus you find out the mole ratio.
mole of manganese metal used = 1.748 / Ar of Mn
= 1.748/ 54.9
= 0.0318 mol
concentration of Mn2+ in stock = 0.0318 mol/dm3
concentration of A = 0.318 x 50/1000/1000
= 0.00000159 mol/dm3
firstly you multiply by 50/1000 to find the amt of mole in 50ml of stock solution. then you divide by 1000 to get concentration of A after dilution.
i hope this helps.
2007-09-29 07:03:04 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The stock coln is of 1.748 g/l conc.
50 ml will have 1/20 * 1.748 g
2007-09-22 12:17:33 · answer #2 · answered by ag_iitkgp 7 · 0⤊ 0⤋
4?
2007-09-29 19:12:43 · answer #3 · answered by OLLIE 4 · 0⤊ 0⤋