Answer:
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
Answer:
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
