What mass of FeCl_3 must be dissolved in enough water to give 555. mL of solutio

Question

What mass of FeCl_3 must be dissolved in enough water to give 555. mL of solution having a final chloride ion concentration of 2.39 M? The molar mass of FeCl_3 is 162.203 g/mol. 2.15 times 10^2 g 48.1 g 71.7 g 18.3 g 61.6 g 637. mL of 0.637 M C_6H_12O_6(aq) was diluted with 599. mL of water. What is the resulting concentration of this solution? 0.677 M 0.391 M 0.328 M 0.387 M 0.361 M What volume, in mL, of 1.44 M Ca(OH)_2(aq) is needed to COMPLETELY NEUTRALIZE 127. mL of 1.29 M HCl(aq)? 56.9 mL 2.28 times 10^2 mL 34.1 mL 90.5 mL

Explanation / Answer

Ans. FeCl3(aq) -----------> Fe3+ + 3Cl-

1 mol FeCl3 dissociates into 1 mol Fe3+ and 3 mol Cl-. Therefore, the solution has 3 times Cl- concentration that of FeCl3 concertation.

Given, [Cl-] = 2.39 M

So, [FeCl3] = (1/3) x [Cl-] = (1/3) x 2.98 M = 0.7966 M

Now,

Moles of FeCl3 = Molarity of FeCl3 x Volume of solution in liters

                                    = 0.7966 M x 0.555 L                                   ; [1 L = 1000 mL]

                                    = 0.442113 mol                                           ; [1 M = 1 mol/ L]

Mass of FeCl3 in solution = Moles x Molar mass

                                    = 0.442113 mol x (162.203 g/mol)

                                    = 71.71 g

Thus, required mass of FeCl3 = 71.71 g

Correct option. C. 71.7 g

#2. Initial volume of solution, V1 = 637 mL

Volume of water added = 599 mL

Final volume of the solution, V2 = 637.0 mL + 599.0 mL = 1236 mL

Initial concertation of glucose = 0.637 M

Now, Using    C1V1 = C2V2                        - equation 1

C1= Concentration, and V1= volume of initial solution 1     ; Glucose original soln.

C2= Concentration, and V2 = Volume of final solution 2          ; Diluted solution

Putting the values in equation 1-

            (637 mL x 0.637 M) = C2 x 1236 mL

            Or, C2 = (637 mL x 0.637 M) /1236 mL = 0.328 M

Thus, final concertation after dilution = 0.328 M

Correct option. C. 0.328 M

#3. Balanced reaction: Ca(OH)2(aq) + 2 HCl(aq) --------> CaCl2(aq) + 2 H2O(aq)

Stoichiometry: 1 mol Ca(OH)2 is neutralized by 2 mol HCl.

Moles of HCl = Molarity x Volume in liters

                        = 1.29 M x 0.127 L

                        = 0.16383 mol

See the stoichiometry, 2 moles of HCl neutralizes 1 mol Ca(OH)2.

Thus,

Moles of Ca(OH)2 required = (1/2) x moles of HCl

                                                = (1/2) x 0.16383 mol

                                                = 0.081915 mol

Let the required volume of Ca(OH)2 be V liters.

So, moles of Ca(OH)2 =Molarity x Volume in liters

            Or, 0.081915 mol = 0.144 M x V

            Or, V = 0.081915 mol /0.144 M = 0.05689 L = 56.89 mL

Thus required volume = 56.89 mL

Correct option: A. 56.9 mL

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