1. Determine the molarity for lead (II) nitrate at 30 degrees Celsius 2. If a so
ID: 1060889 • Letter: 1
Question
1. Determine the molarity for lead (II) nitrate at 30 degrees Celsius 2. If a solution of lead(II) nitrate is prepared by dissolving 150.00 g Pb(NO3)2 in 200.00 ml of water at 50 degrees Celsius, what mass of Pb(NO3)2 will precipitate out of the solution if the solution is cooled to 10 degrees Celsius? 3. Would the decreasing the amount of water from 200.00 ml to 150.00 ml increase or decrease the mass of Pb(NO3)2 that precipitates out at 10 degrees Celsius? Briefly explain your reasoning. 1. Determine the molarity for lead (II) nitrate at 30 degrees Celsius 2. If a solution of lead(II) nitrate is prepared by dissolving 150.00 g Pb(NO3)2 in 200.00 ml of water at 50 degrees Celsius, what mass of Pb(NO3)2 will precipitate out of the solution if the solution is cooled to 10 degrees Celsius? 3. Would the decreasing the amount of water from 200.00 ml to 150.00 ml increase or decrease the mass of Pb(NO3)2 that precipitates out at 10 degrees Celsius? Briefly explain your reasoning. 100T-7-----T 80 70 50 NaCl- 30 03 20 10 Ce2(SO4)- 10 20 30 40 50 60 70 80 90 100 Temperature () 00 90 60 50 40 987654321 (otH 6 001 ul res yo 6) KylignIog
Explanation / Answer
1. amount og Pb(NO3)2 in solution at 30 oC = 65 g/100 ml
molarity of solution = 65/331.2 x 0.1 = 1.962 M
2. Solubility of Pb(NO3)2 at 10 oC = 46 g/100 ml H2O
amount of salt in solution = 150 g/200 ml
So, it is = 75 g/100 ml
When cooled to 10 oC,
amount of Pb(NO3)2 to precipitate out = 75 - 46 = 29 g
3. Decreasing the amount of water from 200 ml to 150 ml would increase the mass of Pb(NO3)2 that prcipitates out at 10 oC as the volume of water now becomes 75 ml instead of 100 ml and it would dissolve thus lesser amount of salt in solution.
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