Written by AH Calculate the following(5 pts each) Molarity of 180 mL of a soluti
ID: 693627 • Letter: W
Question
Written by AH Calculate the following(5 pts each) Molarity of 180 mL of a solution with 17.3 g of phosphoric acid. (5pt) a. 0.980.7M b. 0.980 M c 0.98 M d. 0.981 M 3. Volume of 13.7 M KCIO, containing 15.5 g KCIO, (5pt) a. 1.68 mL b. 9.23 mL c. 2.35 ml d. 2.48 ml 4. 5. Grams of barium hydroxide in 1.5 L of 0.556 M barium hydroxide. (5pt) a. 142.89 g b. 143 g c. 140 g d. 142.9 g Moles of solute in 2.90 L of 0.132 M AgNO, (Spt) a. 0.381 mol H:0 b. 0.381 mol AgNO c. 0.383 mol AgNO d. 0.383 mol Ag 6. 7. A 1.23 g sample of hydrogen peroxide solution generates 42.9 mL O, at 833.1 mmHg pressure and 28°C what is the percent of peroxide in the solution? Partial pressure of H20 at 28'C is 28.3 torr. (10 pts) a. 61.43% H2O2 b. 5.14% H2O2 c. 2.22% H2O2 d. 23.91% H2O1
Explanation / Answer
3)
Molar mass of H3PO4 = 3*MM(H) + 1*MM(P) + 4*MM(O)
= 3*1.008 + 1*30.97 + 4*16.0
= 97.994 g/mol
mass of H3PO4 = 17.3 g
we have below equation to be used:
number of mol of H3PO4,
n = mass of H3PO4/molar mass of H3PO4
=(17.3 g)/(97.994 g/mol)
= 0.1765 mol
volume , V = 180 mL
= 0.18 L
we have below equation to be used:
Molarity,
M = number of mol / volume in L
= 0.1765/0.18
= 0.98 M (because 0.18 has 2 significant figures only)
Answer: c
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