ChemTopper Free Chemistry Help

Here it is
[attachment id=0 msg=2579]
[attachment id=1 msg=2579]
[attachment id=2 msg=2579]

Please post questions one at a time with clear topic

SO2 turns acidified K2Cr2O7 solution to colourless solutions.
Is It true or false
It is true as SO2 reduces  Cr6+ to Cr3+
Chromium shows different color in different oxidation numbers
[attachment id=1 msg=2584] 
It is also important to know that Chromate and dichromate also has different color and chromate (yellow Color)exists in basic condition but changes to dichromate(orange) in acidic condition.
  [attachment id=0 msg=2584]

100 mL of NaHC2O4 required 50 mL of 0.1 M KMnO4 solution in acidic medium. Volume of 0.1 M NaOH required by 100 mL of NaHC2O4 is:

It is an example of double titration
In first titration hydrogen oxalate is oxidized to carbon dioxide and per manganate is reduced to Mn2+
Use the molar ratio of of hydrogen oxalate and potassium permanganate to calculate the moles of sodium hydrogen oxalate from first titration

Two moles of permanganate are reacting with 5 moles of hydrogen oxalate
Now moles of permanganate = molarity X volume in litres= 0.1*50/1000=0.005 moles
Hence moles of hydrogen oxalate = 5* 0.005 /2 moles = 0.025/2 moles
[attachment id=0 msg=2585]
In second titration hydrogen oxalate is behaving like an acid and titration is acid base titration
Here also check the molar ratio of sodium hydroxide and sodium hydrogen oxalate in balanced reaction
In this case one mole of of hydrogen oxalate reacting with one mole of sodium hydroxide
  [attachment id=1 msg=2585]
Use the molar ratio to calculate the moles  of sodium hydroxide and then volume
Hence the moles  sodium hydroxide are the same as that of the hydrogen oxalate in 100 ml

Molarity equals to moles divided by volume in litres
Hence volume = moles / volume in litres
Volume of NaOH = 0.025 /2*0.1=0.125 L = 125mL

In the following titrations indicators are

I : Fe²⁺ vs. Cr₂O₇^2–          II : CI– vs. Ag+          III : Fe²⁺ vs. MnO₄^–

            I                                  II                                     III

(a) self MnO4–,                  [Fe(CN)6]4– ,                        CrO42–

(b) [Fe(CN)6)3– ,                 CrO42– ,                              CNS‾

(c) [Fe(CN)6]3– ,                 CrO42– ,                             self MnO4–   

(d) CNS‾ ,                           CrO42– ,                            [Fe(CN)6]3–



MnO₄^- is a self indicator and changes color with change in oxidation number
MnO₄^--------> Mn ²⁺
    Pink                                                     colorless


[Fe(CN)6]3− + e− ⇌ [Fe(CN)6]4−(purssian blue color)


Both chloride and chromate ions react with silver ions  to give insoluble  colored salts
Ag⁺  + CrO₄ ^2- ---> Ag₂CrO₄ (organge ppt)-Ksp=1.2 x 10^-12
Ag⁺  + Cl^- ----> AgCl (white ppt) Ksp=1. x 10^-10
If molar solublities are calculated then Ag₂CrO₄  has more solubility.
If both ions are present then white silver chloride appears first because of less molar solubility.Once all the chloride ions have reacted, the chromate ions  react and a red -orange precipitate will appear. So appearance of the red-orange silver chromate can be used to indicate the end point of the titration of Cl^- with Ag⁺.

1 molal urea means 1 mole of urea in 1 kg of water
Formula of urea is NH²CONH²

Urea molar mass is 60 gram per mole

1 molal urea means 60 grams of urea is present in 1060 grams of the solution
(1000 + 60 grams of urea)
Now we need to work out that how much is present in in 1000 grams of solution
   1000*60/1060 = 56.6038 g

Question -. 1 g of X % H2O2 required x mL of KMnO4 solution in acidic medium. Thus, molarity of KMnO4solution is:

Select one:
a. 0.60 M
b. none of correct
c. 0.12 M
d. 0.024 M

One gram of x percent hydrogen peroxide means- one gram is impure and only x percent of this 1 gram is pure H₂O₂
It means pure hydrogen peroxide = 1*x/100=0.01 x  gram
Molar mass of hydrogen peroxide is 34 grams
So number of moles in 0.01 x  gram = 0.01  x / 34

Balanced redoxed reaction  of hydrogen peroxide reacting with KMnO₄

Here are balance half reactions
5 H₂O ₂  -----> 5O₂ + 10 e^- + 10H⁺
2MnO₄^- +10e ^- +  16H⁺ ------>2 Mn ²⁺ +  8H₂O 


In this reaction 5 moles of H₂ O₂ are reacting with 2 moles of KMnO₄
number of moles required of KMnO4 = 0.01  x / 34 moles H2O2 * 2 moles KMnO₄/ 5 moles H₂O₂ = 1.2*10^-4 x moles
Total volume of KMnO₄ =xmL= 0.001xL
Molarity is equal to moles divided by volume in litres
Plug in the values to calculate the molarity of KMnO₄
  = 1.2*10^-4 x moles /0.001xL = 0.12 M