The molality of lead nitrate in 0.726 M Pb(NO3)2 is: (The density of the solution is...

Question

Question

The molality of lead nitrate in 0.726 M Pb(NO3)2 is: (The density of the solution is 1.202 g/ml)

science chemistry

Answer 1

Answer #1

moloirty heoterdyU e = 0-126 M density of the solu Han = 1.202 g/ml let volume be l000 m. SOHP:. of lead vithale dansity X mo

Please like this answer. THANK YOU

Please comment if YOU have any doubts about this answer.

All the best for your future.

Answer 2

Similar Homework Help Questions

an aqueous solution is .907M Pb(NO3)2. what is the molality of the lead(II) nitrate in this...

an aqueous solution is .907M Pb(NO3)2. what is the molality of the lead(II) nitrate in this solution? the density of the solution is 1.252g/ml
12. A solution of 0.00016 M lead (II) nitrate, or Pb(NO3)2, was poured into 450 mL...

12. A solution of 0.00016 M lead (II) nitrate, or Pb(NO3)2, was poured into 450 mL of 0.00023 M sodium sulfate, Na2S04. Would a precipitate of lead(II)sulfate, PbSO4, be expected to form if 250 mL of the lead nitrate solution were added? Write the equilibrium equation of PbSO4 dissociation in water.
A solution of NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO3)2(aq) Pb ( NO 3 ) 2 ( aq ) , until no furth...

A solution of NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO3)2(aq) Pb ( NO 3 ) 2 ( aq ) , until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 10.47 g PbCl2(s) 10.47 g PbCl 2 ( s ) is obtained from 200.0 mL 200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq) Pb ( NO 3 ) 2 ( aq ) solution. concentration: M

8. A solution of NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO3)2(aq), until...

8. 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 weighed. A total of 12.12 g PbCl2(s) is obtained from 200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq) solution. concentration: M
A solution of NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO3)2(aq), until no further precipitation occur...

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 weighed. A total of 15.59 g PbCl2(s) is obtained from 200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq) solution. concentration: M
When a solution of Lead Nitrate, Pb(NO3)2 , is mixed with a solution of sodium iodide,...

When a solution of Lead Nitrate, Pb(NO3)2 , is mixed with a solution of sodium iodide, NaI, a precipitate of lead iodide, PbI2 , is formed. a) Write the chemical equation for this reaction showing the state of all reactants and products. b) If 1 mol of Pb(NO3)2 reacts with 2 mols of NaI, what is the amount of PbI2 that will be formed? c) If 1 mol of Pb(NO3)2 reacts with 2 mols of NaI, which compound would be...

Suppose we have a solution of lead nitrate, Pb(NO3)2(aq). A solution of NaCl(aq) is added slowly...

Suppose we have a solution of lead nitrate, Pb(NO3)2(aq). A solution of NaCl(aq) is added slowly until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 15.93 g of PbCl2(s) is obtained from 200.0 mL of the original solution Calculate the molarity of the Pb(NO3)2(aq) solution. Number
A solution of 0.5 M lead(II) nitrate, Pb(NO3)2(aq) is added to an equal volume of 1.0...

A solution of 0.5 M lead(II) nitrate, Pb(NO3)2(aq) is added to an equal volume of 1.0 M sodium iodide, NaI(aq), and lead(II) iodide precipitates, PbI2(s). What is the molar concentration of lead ions, Pb2+(aq), that remains in solution? [FWPbI2 = 461.01 g/mol, Ksp = 1.4 x 10−8]? Assume 298 K.
Potassium iodide reacts with lead (ii) nitrate in the following precipitation reaction: 2KI (aq) + Pb(NO3)2...

Potassium iodide reacts with lead (ii) nitrate in the following precipitation reaction: 2KI (aq) + Pb(NO3)2 (aq)---> 2KNO3 (aq) + PbI2 (s) What minimum volume of 0.200 M potassium iodide solution is required to completely precipitate all the lead in 155.0 mL of a 0.122 M lead (ii) nitrate solution?

A26. What will be observed when 15.0 mL of 0.040 M lead(II) nitrate, Pb(NO3)2, is mixed...

A26. What will be observed when 15.0 mL of 0.040 M lead(II) nitrate, Pb(NO3)2, is mixed with 15.0 mL of 0.040 M sodium chloride? (lead chloride Ksp = 1.7 × 10–5). (A) A clear solution with no precipitate will result. (B) Solid PbCl2 will precipitate and excess Pb2+ ions will remain in solution. (C) Solid PbCl2 will precipitate and excess Cl– ions will remain in solution. (D) Solid PbCl2 will precipitate and there will be no excess ions in solution....