Question

Chemistry: Molarity of Solutions

Directions: Solve each of the following problems. Show your work and include units for full credit.

1. What mass of the following chemicals is needed to make the solutions indicated?

A.. 2.0 liters of a 1.5 M sodium nitrate (NaNO3) solution

B. 5.0 liters of a 0.1 M Ca(OH)2 solution

C. 100 mL of a 0.5 M (NH4)3PO4 solution

2. Calculate the molarity of the following solutions.

A. 198 g of barium bromide (BaBr2) in 2.0 L of solution

B. 54 g of calcium sulfide (CaS) in 3.0 L of solution

3. Calculate the volume of each solution, in liters.  

A. a 0.5 M solution containing 250 g of manganese (II) chloride (MnCl2)

B. a 0.4 M solution containing 290 g of aluminum nitrate (Al(NO3)3)

Answer 1

number of moles of solute Molatry = (n) (m) volume of solubar (4) Memer of moles of solute - Morae mass (A) Molesef of Nanos= = 1.5 mol x 2 tutte 3.0 miles muss - 3,0 mon X 84.998/oster 254.9 a I mars = (0) Molesof pe = caCOH)2 1 Schattex slitex o moj Luf = 0.5 mol Mass = o.smax 74.0929/mor - [ 37.0469] e - Molesef 0.5 mg x (Nhyb poy = Holes of 1oo lite 0.5 mol x d.os.Lute 0.05 mos = Loiolesse CS Scanne peaks 0.05 moix 132.669/anar 26.69 CamScanner

© (A) Moles of Baßrga 1989 = 0.666 mor 297.1481ormon S poracity= 0.666 mol = 10.333M - 2 lebe 72.1.43 glomer - Mla Moraaly? 0.748mor - 10.249 m] Зube & (0) (A) Moles of uniche 1.98 mol . 250 g = 125 drugimon volume of solubara 1.98 mol 0-15 may = 13.97 kite (²) Moles of ALLNO3)3= 1:36 met 2909 = 282.9969/my 1-36 mol volume of solutara Scanned with CamScanner ou me = 13.4 libe helised 0.4ool