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