A solution of NaCl(aq) is added slowly to A solution of lead nitrate, Pb(NO3)2(aq), until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighted. A total of 10.67 g PbCl2(s) is obtained from 200 mL of the original solution. Calculate the molarity of the Pb(NO3)2 (aq) solution.

concentration: ___ M

The se-ci  bonds is more polar  than s-br bond

    pb(so₄)₂this element is lead (iv) sulfate. lead is a multivalent ion. but we know that sulfate ions, so₄²⁻, have a charge of -2. in order to have the ratios match up, we need pb⁴⁺ (two times as many pb⁴⁺  than so₄²⁻).the molar mass can be calculated by finding the mass of one mole of this compound. consider the following calculation,  where amu represents the atomic mass unit. the amu for each atom involved in this compound can be found on a periodic table.  molar mass = amu pb + 2(amu s + 4 * amu o)from a periodic table (with masses rounded to the nearest tenth),      amu pb = 207.2 g    amu s =  32.1 g    amu o = 16.0 gtherefore    molar mass =  207.2 + 2(32.1 + 4*16.0) = 399.4 gthe molar mass is therefore 399.4 grams per mol