all the barium, as BaS04(s), What volume (in mL) of 0.45 M Na2504 solution is ne
ID: 580349 • Letter: A
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all the barium, as BaS04(s), What volume (in mL) of 0.45 M Na2504 solution is needed to precipitate from 17.S mL of 0.25 N Ba(NO3)2 solution? Ba(NO3)2(aq) + Na2SO4(aq) BaSO4(s) + 2 NaNO3(aq) 11.-/1 pointsZumintro6 15.XP.007. The concentration of a sodium hydroxide solution is to be determined. A 30.0 mL sample of 0.115 M solution requires 44.9 mL of the sodium hydroxide solution to reach the point of neutralization. Calculate the molarity of the NaOH solution. 12.-/2 pointsZumintro6 15.P.041 Suppose 1.34 g of FeCls is placed in a 10.0 mL volumetric flask, water is added, the mixture is shaken to dissolve the solid, and then water is added to the calibration mark of the flask. Calculate the molarity of each ion present in the solution. Fe ion Cl ion 13.-/1 pointsZumIntro6 15.P.061. How much water must be added to 468 mL of 0.183 M HCI to produce a 0.142 M solution? (Assume that the volumes are additive.) 14.-/1 pointsZumIntro6 15.P.071. A sample of sodium hydrogen carbonate solid weighing 0.1005 g requires 47.48 mL of a hydrochloric acid solution to react completely. HCl(aq) + NaHCO3(s) NaCl(aq) + H20(/) + CO2(g) Calculate the molarity of the hydrochloric acid solution. 2Explanation / Answer
10. 1 mol Ba(NO3)2 = 1 MOL Na2SO4
no of mol of Ba(NO3)2 = 17.5*0.25 = 4.375 mmol
no of mol of Na2SO4 = 4.375 mmol
volume of Na2SO4 required = 4.375/0.45 = 9.7 ml
11. 1 mol HCl = 1 mol NaOH
no of mol of HCl = 30*0.115 = 3.45 mmol
no of mol of NaOH = 3.45 mmol
molarity of NaOH = n/M = 3.45/44.9 = 0.077 M
12.
no of mol of FeCl3 = 1.34/162.2 = 0.0083 mol
molarity(M)of FeCl3 = n/V*1000
= 0.0083/10*1000
= 0.83 M
molarity of Fe^3+(aq) = 0.83 M
molarity of Cl^-(aq) = 0.83*3 = 2.49 M
13. M1V1 = M2V2
(0.183*468) = (0.142*V2)
v2 = 603.13 ML
Volume of water added = 603.13 - 468 = 135.13 ml
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