The initial concentration of solution is 0.0693 M. The volume of solution taken is 10 mL and it is diluted to a final volume of 500 mL.
According to dilution law, the product of initial concentration and volume is equal to the product of final concentration and volume as follows:
Here, 
is initial concentration, 
is final concentration, 
is initial volume and 
is final volume.
Rearranging to calculate final concentration,
Putting the values,
Therefore, concentration of the resulting solution is 0.001386 M.
Gets exited and moves at a faster constant speed
1. Cu + 2AgNO3 ==> 2Ag + Cu(NO3)2 ... balanced equation. This is both a single replacement reaction and an oxidation reduction reaction.
Moles of Cu present = 19.0 g Cu x 1 mole Cu/63.55 g = 0.2990 moles Cu
Moles AgNO3 = 125 g AgNO3 x 1 mole AgNO3/169.9 g = 0.7357 moles AgNO3
Which reactant is limiting? It will be Cu because the mole ratio is 2 AgNO3 to 1 Cu and there is more than enough AgNO3. Thus, amount of Ag formed will depend on moles of Cu (0.2990)
Moles of Ag formed = 0.2990 moles Cu x 2 moles Ag/mole Cu = 0.598 moles Ag
Mass (grams) of Ag formed = 0.598 moles Ag x 107.9 g/mole = 64.52 g = 64.5 g of Ag (3 sig. figs.)
A chemist must prepare 0.200 L of 1.00 M aqueous silver nitrate working solution. He'll do this by pouring out 1.82 mol/L aqueous silver nitrate stock solution into a graduated cylinder and diluting it with distilled water. How many mL of the silver nitrate stock solution should the chemist pour out?
Answer: 0.110 L
Explanation:
According to the dilution law,
where,
= molarity of stock silver nitrate solution = 1.82 M
= volume of stock silver nitrate solution = ?
= molarity of diluted silver nitrate solution = 1.00 M
= volume of diluted silver nitrate solution = 0.200 L
Putting in the values we get:
Therefore, volume of silver nitrate stock solution required is 0.110 L
