Question

4. What volume (in L) of 0.586 M Ba(OH)2 (aq) contams V.* of Ba(OH)2 dissolved in it? 5. If 16.0 mL of water are added to 31.5 mL of 0.586 M Ba(OH)2 (aq), what is the new solution molarity?

Answer 1

4. 0.132 L

5. 0.389 M

- 4. [:: 1 ounce = 28-357] mass of Ba(OH), = 0.4-66 ounces- 13:210878 , Moles of Ba(OH), in 13:2108789 given mass of Ba(OH)2 Mular mass of Ba(OH)2 = 13.2108789 171:34 g/mol = 0.0171032917 moles. SWOLD - Required concentration = 0.586 M fon so, Molarity, M = mules volume l Litre) 0.586 = 0.0771032917 volume (litre) Volume (litre) = 0.077.1032917 0.586 = 0.1315755 831 L er, Volume = 0.132 L 5. Males of Ba(OH), in 315 ml 0.586 M Ba(OH)2 1000 ml contains I ml Solution contain - 0.586 0.586 moles of Ba(OH)2 moles, ODD 31.5 me solution contain - 0.586 X 31-5 1000 -0.018459 moles of Ba(OH)2 Final volume of solution = (V+V2) =( 31.5 ml + 160 me) = 47.5 mL 0.0475L Molarity = moles - Volume (litre)

Molarity, M = 0.018459 0.0475 = 0.3886105263 M 0. 389 M ar Molarity