H2O2 acts as both oxidising as well as reducing agent its product is H2O, when act as oxidising agent and its product is 02 when act as of reducing agent. The strength of '10P means one liter of H2O2 solution on decomposition at S.T.P. condition liberate 10 liters of oxygen gas (H2O → H2O + 1/2 O2) 15 gm Mn04)2 sample containing inert impurity is completely reacting with 100 ml of '11.2 V' H2O2, then what will be the % purity of Ba (Mn04)2 in the sample? (Atomic mass Ba = 137, Mn = 55)
1.2 gm of carbon is burnt completely in oxygen (limited supply) to produce CO and CO2. This mixture of gases is treated with solid I2O5 (to know the amount of CO produced), the liberated iodine required 120 ml of 0.1 M hypo solution for complete titration. The % of carbon converted into CO is:
Hence, the correct answer is Option D.
Nitrogen (N), phosphorus (P) and Potassium (K) are the main nutrients in plant fertilizers. According to an industry convention, the numbers on the label refer to the mass % of N, P2O5 and K2O in that order calculate the N : P : K ratio of a 30 : 10 : 10, fertilizer in terms of moles of each element and express it as x: y : 1.
Molar mass of N is 14.0067 g/mol.
Molar mass of P2O5 is 30.973762*205 = 6349.62121 g/mol.
Molar mass of K2O is 39.0983*20 = 781.966 g/mol
Ratio of masses is 30:10:10
Therefore for each 50 g of fertilizer:
For N: 30 g / 14.0067 g/mol = 30/14.0067 moles
For P2O5: 10 g / 6349.62121 g/mol = 10/6349.62121 moles
For K2O: 10 g / 781.966 g/mol = 10/781.966 moles
Ratio of elements in terms of moles:
30/14.0067 : 10/6349.62121 : 10/781.966
30/14.0067*781.966/10 : 10/6349.62121*781.966/10 : 10/781.966*781.966/10
167.484 : 0.123 : 1.0
In what ratio should a 15% solution of acetic acid be mixed with a 3% solution of the acid to prepare a 10% solution(all percentages are mass/mass percentages):
105 ml of pure water at 4°C saturated with NH3 gas yielded a solution of density 0.9 g/ml and containing 30% NH by mass. Find the volume of resulting NH3 solution.
volume of water = 105 mL
density of solution = 0.9 g/mL
Since ammonia dissolved in water is 30% by weight, then using mass % formula, we can calculate the amount of ammonia in water
At 4 degree C, density of water = 0.999 g/mL
since density = mass /volume
0.999 = mass/ 105
mass of water = 104.895 g
now, let mass of ammonia be x.
30x + 3146.85 = 100x
70x = 3146.85
x = 44.955 g
Thus, ammonia dissolved in 105 mL water is 44.955 g
Therefore, total mass of solution = mass of ammonia + mass of water
Mass of solution = 44.955 + 104.895
Mass of solution = 149.85 g
So,density of solution = mass of solution / volume of solution
0.9 g/mL = 149.85 g / Volume of solution
Volume of solution = 149.85/ 0.9
Volume of solution = 166.5 mL
The equivalent mass of H3BO3 (M = Molar mass of H3BO3) in its reaction with NaOH to from Na2B4O7 is equal to –
X gram of pure As2S3 is completely oxidised to respective highest oxidation states by 50 ml of 0.1 M hot acidified KMnO4 then x mass of As2S3 taken is : (Molar mass of As2S3 = 246)
The number of moles of ferrous oxalate oxidised by one mole of KMnO4 is
How many moles of KMnO4 are needed a mixture of 1 mole of each FeSO4 & FeC2O4 in acidic medium
In the reaction
Na2S2O3 + 4C12 + 5H2O → Na2SO4 + H2S04 + 8HCI
the equivalent weight of Na2 S2 O3 will be
(M= molecular weight of Na2S2O3)
C2H4 + xO2 -> 2CO2 + yH2O, what is the value of x + y?
The balance combustion equation is C2H4 + 3O2 -> 2CO2 + 2H2O, => x = 3, y = 2, => x + y = 5.
The following equations are balanced atomwise and chargewise.
(i) Cr2O72- + 8H+ + 2H2O2 → 2Cr3+ + 7H2O + 2O2
(ii) Cr2O72- + 8H+ + 5H2O2→ 2Cr3+ + 9H2O + 4O2
(iii) Cr2O72- + 8H+ + 7H2O2→ 2Cr+ + 11H2O + 5O2
The precise equationl equations representing the oxidation of H2O2 is/are
The correct answer is option A
Cr2O72- converts into Cr3+ in acidic medium I.e. in H+ medium.
First balance the Cr atom on both sides and then Oxygen atom. H+ is in excess due to acidic medium.
Add H+ as +ve charge to balance the charge on both sides.
An excess of NaOH was added to 100 mL of a ferric chloride solution. This caused the precipitation of 1,425 g of Fe(OH)3. Calculate the normality of the ferric chloride solution
In the reaction CrO5 + H2SO4 →Cr2(SO4)3 + H2O + O2 one mole of CrO5 will liberate how many moles of O2
The balanced chemical reaction is,
From the balanced chemical reaction, we conclude that
As, 4 moles of CrO5 react to give 7 moles of O2
So, 1 mole of CrO5 react to give 7/4 moles of O2
0.4g of a polybasic acid HnA (all the hydrogens are acidic) requires 0.5g of NaOH for complete neutralization. The number of replaceable hydrogen atoms in the acid and the molecular weight of 'A' would be : (Molecular weight of the acid is 96 gms.)
A solution of Na2S2O3 is standardized iodimetrically against 0.1262 g of Mr% This process requires 45 mL of the Na2 S203 solution. What is the strength of the Na2S2O3?
25.0 g of FeSO4.7H2O was dissolved in water containing dilute H2SO4, and the volume was made up to 1.0 L. 25.0 mL of this solution required 20 mL of an N/10 KMnO4 solution for complete oxidation. The percentage of FeSO4. 7H2O in the acid solution is
1.0 mol of Fe reacts completely with 0.65 mol of O2 to give a mixture of only Fe0 and Fe2O3. The mole ratio of ferrous oxide to ferric oxide is
2Fe + O2 →→ 2FeO
4Fe + 3O2 →→2Fe2O3
Let x moles of Fe reacts to produce FeO
Thus amount of Fe reacted to produce Fe2O3 = 1-x Moles
2 Moles of Fe reacts to produce FeO with = 1 moles of O2
1 moles of Fe reacts to produce FeO with = 1/2 moles of O2
x moles of Fe to produce FeO with = 0.5x moles of O2
4 moles of Fe reacts to produce Fe2O3 with = 3 moles of O2
1-x moles of Fe reacts to produce Fe2O3 = (3/4) x (1-x) Moles of O2
1) 0.5x + 3/4(1−x) =0.65 moles of O2
x = 0.4 moles
2 Moles of Fe produces = 2 mole of FeO
0.4 moles of Fe produces = 0.4 moles of FeO
4 moles of Fe produces = 2 moles of Fe2O3
(1-0.4 )moles of Fe produces = (1/2) x 0.6 Moles = 0.3 moles of Fe2O3
FeO/Fe2O3 = 0.4/0.3 = 4/3
25 mL of a solution containing HCf and H2S04 required 10 mL of a 1 N NaOH solution for neutralization. 20 mL of the same acid mixture on being treated with an excess of AgNO3 gives 0.1425 g of AgCf. The normality of the HC and the normality of the H2S04 are respectively
If 10 gm of V2O5 is dissolved in acid and is reduced to V2+ by zinc metal, how many mole of I2 could be reduced by the resulting solution if it is further oxidised to VO2+ ions ?
[Assume no change in state of Zn2+ions] (V = 51, 0 = 16, I = 127) :
0.70 g of mixture (NH4)2 SO4 was boiled with 100 mL of 0.2 N NaOH solution till all the NH3(g) evolved and get dissolved in solution itself. The remaining solution was diluted to 250 mL. 25 mL of this solution was neutralized using 10 mL of a 0.1 N H2SO4 solution. The percentage purity of the (NH4)2 SO sample is
A mixed solution of potassium hydroxide and sodium carbonate required 15 mL. of an N/20 HCI solution when titrated with phenolphthalein as an indicator. But the same amount of the solution when titrated with methyl orange as an indicator required 25 mL of the same acid. The amount of KOH present in the solution is
The percentage of copper in a copper(II) salt can be determined by using a thiosulphate titration. 0.305 gm of a copper(II) salt was dissolved in water and added to, an excess of potassium iodide solution liberating iodine according to the following equation
2Cu2 (aq) + 4I– (aq) 2CuI(s) + I2(aq)
The iodine liberated required 24.5cm3 of a 0.100 mole dm-3 solution of sodium thiosulphate
2S2032- (aq) + I2(aq) → 2I– (aq) + S4062- (aq)
the percentage of copper, by mass in the copper (ll) salt is. [Atomic mass of copper = 63.5]
KIO3 reacts with KI to liberate iodine and liberated Iodine is titrated with standard hypo solution, The reactions are
1. IO3– + I– →I2 (valency factor = 5/3)
2. I2 + S2O32– → S4O62– + I– (valency factor = 2)
meq of hypo = meq of I2 = meq of IO3– + meq of I–
IO3– react with I– ⇒ meq of IO3– = meq of I–
Statement-1 : meq of hypo = 2 x meq of IO3–
Statement-2 : valency factor of I2 in both the equation are different therefore we cannot equate milliequivalents in sequence
The correct answer is option A
Milliequivalent of IO3− is equal to milliequivalent of I−.
Milliequivalent of hypo is equal to milliequivalent of IO3− plus milliequivalent of I−.
The milliequivalent of hypo is twice the milliequivalent of IO3−. It is also twice the milliequivalent of I−.
In the reaction:
2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
how many moles of CO2 are formed when 1 mole of O2 is consumed?
1 mole O2 x (4 mole CO2 / 7 mole O2 ) = 4/7 mole CO2