You need to make an aqueous solution of 0.196 M lead nitrate for an experiment in lab, using a 125 mL volumetric flask. How much solid lead nitrate should you add

We have to ad 8.11 g of Pb(NO₃)₂

Explanation:

To determine the mass of solute that would be needed for this experiment we have to think in the relation of Molarity and volume

Molarity = Mol of solute / Volume (L)

Therefore, Molarity . Volume (L) = Mol of solute

We convert the volume of the volumetric flask in mL to L

125 mL . 1L / 1000mL = 0.125 L

0.196 mol/L . 0.125L = Moles of solute → 0.0245 moles

Now we convert these moles, to mass

Solute: Pb(NO₃)₂ → Molar mass: 331.2 g/mol

Moles . molar mass = mass → 0.0245 mol . 331.2 g/mol = 8.11 g

We have to add 8.11 grams of lead nitrate

Explanation:

Step 1: data given

Molarity of lead nitrate (Pb(NO3)2) = 0.196 M

Volume = 125 mL = 0.125 L

Molar mass Pb(NO3)2 = 331.2 g/mol

Step 2: Calculate moles Pb(NO3)2

Moles Pb(NO3)2 = volume Pb(NO3)2 * molarity Pb(NO3)2

Moles Pb(NO3)2 = 0.125 L * 0.196 M

Moles Pb(NO3)2 = 0.0245 moles

Step 3: Calculate mass Pb(NO3)2

Mass Pb(NO3)2 = moles Pb(NO3)2 * molar mass Pb(NO3)2

Mass Pb(NO3)2 = 0.0245 moles ¨331.2 g/mol

Mass Pb(NO3)2 = 8.11 grams

We have to add 8.11 grams of lead nitrate

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error

i think it's letter a.