11. What is the molarity of a solution prepared by dissolving of water? $4.3 g o
ID: 591554 • Letter: 1
Question
11. What is the molarity of a solution prepared by dissolving of water? $4.3 g of Ca(NO,), into 335 mL A) 0.331 AM B) 0.932 M C) 0.117 M D) 1.99 M E) none of the above 12. A 0.15 M solution of BaCl2 contains: A) 0.15 M Ba2+ ions and 0.15 M Cl- ions. B) 0.15 M Ba2+ ions and 0.30 M CI- ions. C) 0.30 M Ba2+ ions and 0.15 M CI-ions. D) 0.30 M Ba2+ ions and 0.30 M Cl- ions. E) none of the above 13. What is the final concentration of a solution prepared by diluting 35.0 mL of 12.0 M HCI to a final volume of1.20L? A) 0.504 M B) 3.50 M C) 0.420 M D)0.350 M E) none of the above 14. Which of the following statements about colligative properties is FALSE? A) The boiling point of a solution is increased by the addition of salt. B) The freezing point of a solution is lowered by the addition of salt. C) The change in temperature is proportional to the molality. D) The identity of the solute is not a factor. E) All of the above statements are true. 15. Which of the following solutions would have the highest boiling point? A) 1.0 M KBr B) 1.0 M C6H1206 C)1.0 M CaCl2 D) Each of these solutions will have the same boiling point. E) not enough information 16. What is the molality of a solution made by dissolving 14.7 g of C,Hn0, into 150.0 mL of water? Assume the density of water is1.00 g/ml. Public InformationExplanation / Answer
11)
Molar mass of Ca(NO3)2 = 1*MM(Ca) + 2*MM(N) + 6*MM(O)
= 1*40.08 + 2*14.01 + 6*16.0
= 164.1 g/mol
mass of Ca(NO3)2 = 54.3 g
we have below equation to be used:
number of mol of Ca(NO3)2,
n = mass of Ca(NO3)2/molar mass of Ca(NO3)2
=(54.3 g)/(164.1 g/mol)
= 0.3309 mol
volume , V = 335 mL
= 0.335 L
we have below equation to be used:
Molarity,
M = number of mol / volume in L
= 0.3309/0.335
= 0.988 M
Answer: 0.988 M
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