posted by Mohammed .
A saturated solution of lead ii trioxonitrte v was prepared at 22degree Celsius at 27cm3 of this solution required 46cm3 of NaCl solution containing 96gdm3 for complete precipitation. Find the solubility of the lead in trioxonitrte v at 22degree Celsius in (a) mol/dm3 (b) g/dm3.
Pb(NO3)2 + 2NaCl ==> PbCl2 + 2NaNO
mols NaCl = grams/molar mass = 96/58.45 = approx 1.6 but you should confirm this as well as all other calculations that follow. I've estimated.
1.6 mols/dm3 = 1.6 M
mmols NaCl = mL x M = 46 x 1.6 = approx 75.
That means the amount of Pb(NO3)2 in solution is 1/2 that (look at the coefficients in the balanced equation) or 75/2 = approx 37.
mols Pb(NO3)2 = 0.037
27 cc = 0.027 cm^3
M Pb(NO3)2 = mols/dm^3 = ?
for part b convert. Im sure you can do that.