Answer to Question #179183 in Chemistry for Alex

Question

how many grams of lead nitrate is needed to make 450 mL solution with the concentration of 1.25 mol/L

Answer 1

Solution:

Lead(II) nitrate - Pb(NO₃)₂

The molar mass of Pb(NO₃)₂ is 331.2 g/mol.

Hence,

(0.45 L) × (1.25 mol Pb(NO₃)₂/ 1 L) × (331.2 g Pb(NO₃)₂ / 1 mol Pb(NO₃)₂) = 186.3 g Pb(NO₃)₂

Answer: 186.3 grams of lead nitrate (Pb(NO₃)₂) is needed.