1. What is the amount of oxalate in 2 mL of 0.035M solution?
a) 7 ¡Á 10¨C2 mole
b) 1.75 ¡Á 10¨C2 mole
c) 7 ¡Á 10¨C5 mole
d) 1.75 ¡Á 10¨C5 mole
2. Sodium oxalate may be prepared by titrating oxalic acid (C2H2O4) with
sodium hydroxide, according to the equation:
C2H2O4(aq) + 2NaOH(aq) ¡ú Na2C2O4(aq) + 2H2O(l)
If the titration of 2 mL oxalic acid with 0.05M NaOH requires 4.3 mL of
titrant, what is the concentration of C2H2O4(aq)?
a) 0.054M
b) 0.043M
c) 0.22M
d) 0.11M
3. The concentration of FeSO4.7H2O (MW = 278.0) is 300 g L¨C1. What is the
concentration in mol L¨C1?
a) 2.78M
b) 0.93M
c) 1.08M
d) 3.00M
4. What are the oxidation numbers of sodium and chlorine in NaCl?
Na Cl
a) 0 0
b) +1 +1
c) +1 ¨C1
d) ¨C1 ¨C1
5. What is the indicator in the titration of hypochlorite with thiosulfate?
a) thiosulfate
b) iodide
c) hypochlorite
d) iodine
5 = a
5= b
5b
1. To find the amount of oxalate in 2 mL of 0.035M solution, you can use the formula:
Amount of substance (in moles) = concentration (in M) * volume (in L)
In this case, the concentration is 0.035M and the volume is 2 mL converted to liters (2 mL = 0.002 L). Plugging these values into the formula:
Amount of substance = 0.035M * 0.002L = 0.00007 moles
So the amount of oxalate in 2 mL of 0.035M solution is 0.00007 moles.
Option a) 7 × 10⁻² mole is incorrect because it's a much larger amount.
Option b) 1.75 × 10⁻² mole is incorrect because it's also a larger amount.
Option c) 7 × 10⁻⁵ mole is incorrect because it's an even smaller amount.
Option d) 1.75 × 10⁻⁵ mole is the correct answer.
2. To find the concentration of C2H2O4(aq) in the given titration, you need to use the concept of stoichiometry and the volume of titrant required.
The balanced equation tells us that 1 mole of C2H2O4 reacts with 2 moles of NaOH. So, the ratio of their concentrations is 1:2.
Given that it took 4.3 mL of 0.05M NaOH to titrate 2 mL of oxalic acid, we can calculate the moles of NaOH used:
Moles of NaOH = concentration (in M) * volume (in L)
Moles of NaOH = 0.05M * 0.0043L = 0.000215 moles
Since the ratio of C2H2O4 to NaOH is 1:2, the moles of C2H2O4 must be half of the moles of NaOH used:
Moles of C2H2O4 = 0.000215 moles * 0.5 = 0.0001075 moles
Now, we can find the concentration of C2H2O4:
Concentration of C2H2O4 = moles of C2H2O4/volume (in L)
Concentration of C2H2O4 = 0.0001075 moles/0.002L = 0.05375 M
Rounded to three significant figures, the concentration of C2H2O4(aq) is 0.054M.
Option a) 0.054M is the correct answer.
3. To convert the concentration of FeSO4·7H2O from g/L to mol/L, you need to use the formula:
Concentration (in mol/L) = mass (in g)/molar mass (in g/mol)
Given that the concentration is 300 g/L and the molar mass of FeSO4·7H2O is 278.0 g/mol, plugging in these values into the formula:
Concentration (in mol/L) = 300 g/L / 278.0 g/mol = 1.0791 mol/L
Rounded to three significant figures, the concentration of FeSO4·7H2O is 1.08M.
Option c) 1.08M is the correct answer.
4. The oxidation number of an element is a number assigned to each atom in a compound. In NaCl, sodium (Na) is a Group 1 element, which means it has a +1 oxidation number. Chlorine (Cl) is a Group 17 element, which means it has a -1 oxidation number.
Therefore, the oxidation numbers of sodium and chlorine in NaCl are:
Na: +1
Cl: -1
Option b) +1 -1 is the correct answer.
5. In the titration of hypochlorite with thiosulfate, the indicator used is typically iodide (I⁻).
During the titration, the reaction between the hypochlorite ions (ClO⁻) and iodide ions (I⁻) produces iodine (I₂). The iodine formed is then titrated with thiosulfate (S₂O₃²⁻) until all the iodine is converted back to iodide.
Therefore, the indicator in the titration of hypochlorite with thiosulfate is iodide.
Option b) iodide is the correct answer.