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.

  1. DrBob222

    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.

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