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A solution of NaCl(aq)NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO3)2(aq)Pb(NO3)2(aq) , until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 13.25 g PbCl2(s)13.25 g PbCl2(s) is obtained from 200.0 mL200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq)Pb(NO3)2(aq) solution.

Respuesta :

Answer:

The molarity of the Pb(NO3)2 solution is 0.24 M

Explanation:

Assuming there was no loss during filtration, we can calculated amount of moles Pb(NO3)2 from the 13.25 g of PbCl2 ussing molar mass of  PbCl2

13.25 g / 278.1 g = 0.04 moles of PbCl2

According balanced equation 2NaCl + Pb(NO3)2 = 2NaNO3 + PbCl2, 1 mol of Pb(NO3)2 produces 1 mol of PbCl2. So, 0.04 moles of Pb(NO3)2 produces PbCl2

Molarity is = moles of solute/volumen of solution (L)=0.04/0.2= 0.24

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