Question

Question 3 1.2 pts When 0.385 M RbOH solution is mixed with an aqueous solution of Pb(NO3)2, 15.656 g of precipitate is formed (precipitate should be identified first - see question #1). The number of moles of precipitate formed is The number of moles of RbOH that has reacted is and the volume of the 0.385 M ROOH solution mixed is (MM (g/mole) RbOH = 102.478; Pb(OH)2 = 241.216; Pb(NO3)2=331.21; RbNO3=147.47) (Express your answer with 4 decimal figures or 4 prd's and do NOT include the units). (Hint: this problem is similar to textbook example 4.3 done in class)

Answer 1

* The balanced Chemical equation 2 RbOH ans + Pb(NO), — is > 2 Rb ND + Pb(OH)et (99) (s) coop) (29) 199) As per question given [RBOH] = 0.385 M amount of Polon), formed = 15.656 g # moler mass of Poloh), - 207.2 + 2x (15.999 + 1.008) - 241-214 g/mol number of mole u Precipitate produced 15.656 4 241.214 g/mol = 0.0649 mol * From the balanced chemical equction Imol of Poloh), is formed 'ften when 2 mol RbOH is reacted. → for formation : 0.0 649 molar Ph formed & = 2 x 0.0649 mol op Rbot must be reached no. of moles of Rb oh that has reached : '= 2x0.0649 = 0.1298 mol Opunt

concentration of RbOH = 0.385 M .! Solution contains 0-386 mol in 1 liter or solution Solution mixed for 0.1298 mol of RbOH amount of will be 0.1398 mol 0.385 moll liter = 0.3371 L