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Procedure: (1) Reaction between sodium iodide and lead (II) nitrate: Note the appearance 1. Weigh approximately 3.55 g of Nal (0.0237) and place it in the 100 beaker. Record the exact mass in the data table below. Note the a of the Nal crystals in the observation section. 2. Add 30 mL of distilled water to the beaker containing the Nal crystals. S the solution until all of the Nal crystals have dissolved. Record any observations in the observation section. 3. Weigh approximately 6.55 g of Pb(NO3)2 (0.0198) and place it in the 100-1 beaker. Record the exact mass in the data table below. Note the appearance of the Pb(NO3)2 crystals in the observation section. 4. Add 30 mL of distilled water to the beaker containing the Pb(NO3)2 crystals! Stir the solution until all of the Pb(NO3)2 crystals have dissolved. Record any observations in the observation section 5. Slowly, pour sodium iodide solution into lead (II) nitrate solution. Stir the reaction occasionally. Record your observations of the resulting reaction The reaction should produce "yellow solid" by precipitation process. 6. Set up the Vacuum filtration process as shown below and follow the procedure: 0 Obtain a piece of filter paper and determine its mass. Record the mass in the data table. (11) Pour the reaction mixture into the funnel to separate the solid from the liquid. Make sure all of the solid is collected in the filter paper. Use a distilled water bottle to rinse the solid from the reaction (it) Pour the liquid collected into the waste beaker. Allow the filter paper to dry until the next laboratory period () Obtain the mass of the filter paper and solid. Record theme if necessary. the solid product should be dried in the oven. next laboratory period. Mostly, data table below. The mass of solid product is the actual ord the mass in the Observations and Data Table: Nal crystals Nal Solutions PANO crystals PONO) Solutions Chemical reaction between Nal and Pb(NO3). N rocks Icleor IClear Write the balanced equation: PONO3)2 +2 No IZNO NO₃ + Pb Iz Hallow, clumply o w *

Answer 1

According to the procedure

the mass of NaI is 3.55 g

Number of moles of NaI can be calculated from the molar mass of NaI

Molar mass of NaI = 149.89 g/mol

Number of moles of NaI = (Given weight of NaI)/(Molar mass of NaI) = 3.55 g / 149.89 g/mol = 0.023684 moles

Moles of NaI = 0.023684

Given that

Mass of lead nitrate, Pb(NO₃)₂ = 6.55 gms

Number of moles of lead nitrate can be calculated as in the case of NaI

Molar mass of lead nitrate = 331.2 g/mol

Number of moles of lead nitrate = (Given weight of lead nitrate)/(Molar mass of lead nitrate) = 6.55 g/ 331.2 g/mol = 0.01978

Moles of Pb(NO₃)₂ = 0.01978

Given that NaI is dissolved in 30 mL of water

Molarity is the number of moles of solute present in 1 ltr of solution which can be given by

Molarity = Number of moles x (1000 / Volume of solution in mL)

We have number of moles of NaI = 0.023684

Volume of NaI solution = 30 mL

Therefore Molarity = 0.023684 x (1000 / 30) = 0.7895 mol/L

Molarity of NaI = 0.7895 mol/L

Given that Pb(NO₃)₂ was dissolved in 30 mL of water

Likewise in the above step,

We have number of moles of Pb(NO₃)₂ = 0.01978

Volume of Pb(NO₃)₂ solution = 30 mL

Therefore the molarity = 0.01978 x (1000 / 30) = 0.6593 mol/L

Molarity of Pb(NO₃)₂ = 0.6593 mol/L