Question

62. Determine the molarity of a solution that was prepared using 32.8 g NaCI and enough water to make a 450.0 ml. solution. The molar mass of NaCI is 58.44 g/mol. E e olo 63. How many moles of HCI are in 300.0mL of a 12.0 M HCI solution? A0 64. A 20.0mL sample of HCI is titrated to the endpoint with 35.0mL of a 0.50M NaOH solution. What is the concentration of HC1? paivollosn wollot la noesup otunp i aiwan onalin o snimd r tndslin suim h0 di ayim ED to 0 65. Which of the following aqueous solutions contains the most number of ions? a) 0.10 M LiF b) 0.05 M NaCI c) 0.10 M Mg(NOs)2 ion d) 0.50 M CO2 e) 0.05 M (NH4)sPO4

Answer 1

Answer:

62)

Given : Mass of NaCl = 32.8 g, Molar mass of NaCl = 58.44 g.mol^-1, Volume of solution to be made = 450.0 mL = 0.450 L.

Let us calculate the # of moles of NaCl.

# of moles of NaCl = Mass of NaCl given / Molar mass of NaCl = 32.8 g / 58.44 g.mol^-1 = 0.592 mol.

[NaCl] = # of moles NaCl /Volume of solution in L = 0.592 mol / 0.450 L = 1.32 mol/L = 1.32 L

63)

Given: Molarity of HCl = 12.0 M, Volume of HCl solution = 300.0 mL = 0.3 L

Then,

# of moles of HCl = Molarity of HCl * Volume of HCl in L = 12.0 * 0.3 = 3.6 mol.

# of moles of HCl = 3.6 mol.

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