Respuesta :
The concentration of the stock solution is 0.029 M
The concentration of the solution A is 0.00145 M.
0.000058 moles/litre is the concentration of Solution B.
0.00000116 moles/ litre is the concentration of solution C.
Explanation:
Weight of the manganese metal dissolved is 1.584 and diluted in 1000 ml.
The number of moles of manganese will be calculated as
Number of moles = mass ÷ atomic mass (atomic mass of manganese = 54.93 gram/mole)
Thus number of moles = 1.584 ÷ 54.93
= 0.029 moles
Molarity or concentration of the Mn ions is calculated by the formula:
M= n ÷ V
M = 0.029 ÷ 1
0.029M is the molarity of the solution given.
Now to know the molarity of solution A
It can be known by
M1V1 = M2V2
0.029 × 50 ml = M2 × 1000 ml
M2= 0.00145 M thus the molarity of the solution A is 0.00145 M.
Molarity of solution B ( 10 ml of solution A is diluted to 250 ml)
Applying the formula:
M1V1 = M2V2
0.00145 × 10 = M2 × 250 ( ml will be converted to L by dividing the volume with 1000)
M2 = 0.000058 moles/litre is the concentration of Solution B.
For solution C ( 10 ml of solution B is diluted to 500ml)
From the formula:
M2 V2 = M3V3
0.000058 × 10 = M3 × 500 ( Volume will be changed to L )
0.000058 × 0.01 = M3 × 0.5
= 0.00000116 moles/ litre is the concentration of solution C.
