Question
An excess of silver nitrate solution was added to 10.0cm3of sodium chloride solution, and 0.717g of silver chloride was precipitated. Calculate the concentration of the sodium chloride solution in mol dm-3.
Answers
how many moles of AgCl2 in 0.717g ?
0.717/143.32 = 0.005 moles
Ag(NO3)2 + 2NaCl = AgCl2 + 2NaNO3
So you had 2*0.005 = 0.01 moles of NaCl
0.01mol/0.01L = 1mole/L = 1M NaCl
0.717/143.32 = 0.005 moles
Ag(NO3)2 + 2NaCl = AgCl2 + 2NaNO3
So you had 2*0.005 = 0.01 moles of NaCl
0.01mol/0.01L = 1mole/L = 1M NaCl
Related Questions
A 6.70 g sample of a binary mixture of silver (I) nitrate and sodium nitrate is completely
dissolve...
Hydrolysis of a compound A, C7H6cl2 was carried out by refluxing with excess potassium hydroxide sol...
Identify the precipitate(s) of the reaction that occurs when a silver nitrate solution is mixed with...