4.4g of CO2 and 2.24 litre of H2 at STP (273.15 K and 1 atm pressure) are mixed in a container. The total of molecules present in the container will be
Number of moles of CO2 =
∵ 22.4 L of H2 has 1 mole of H2
∴ The number of moles in 2.24L
The total moles are 0.2.
⇒ Total molecules in 0.2 moles
4I– + Hg2+ → HgI42– ; 1 mole of each Hg2+ and I– will form:
Balanced reaction is as follows :
Hg2+(aq) + 4I- (aq) ⇋ HgI42- (aq)
∵ 4 mole of I- produces = 1 mole of HgI42-
∴ 1 mole I- produces = 1/4 moles of HgI42-
How many grams of I2 are present in a solution which requires 40 ml of 0.11N Na2S2O3 to react with it via the reaction:
S2O32– + I2 → S4O62– + 2I–?
Meq. of I2 = Meq. of Na2S2O3 = 40 × 0.11
∴ (W × 2 × 1000)/254 = 40 × 0.11
W of I2 = 0.558g
Two elements A (atomic mass = 75) and B (atomic mass = 16) combine to yield a compound. The % by weight of A in the compound was found to be 75.08. The formula of compound is:
Molecular weight of A2B3 = 75 × 2 + 16 × 3 = 198
198g of A2B3 =150g of A
∴ 10g of A2B3 = = 75.08g of A
So option B is correct.
What weight of HNO3 is needed to convert 5g of Iodine into Iodic acid according to the reaction:
I2 + HNO3→ HIO3 + NO2 +H2O
The balanced reaction is:
I2 + 10HNO3 → 2HIO3 + 10NO2 + 4H2O
(127 * 2)g of iodine requires (63 * 10)g of HNO3
Thus, 5g of I2 requires
= [(63 * 10)/(127 * 2)] * 5 = 12.4g
In the reaction, VO + Fe2O3 → FeO + V2O5, the equilvalent weight of V2O5 is equal to its
When BrO3– ion reacts with Br– ion in acid solution Br2 is liberated. The eq. weight of KBrO3 in this reaction is:
10e + 2Br5+ → Br02
The hydrated salt, Na2SO4.nH2O, undergoes 55.9 % loss in weight on heating and become anhydrous. The value of n will be:
Molecular Mass of Na2SO4 is = 2 x 23 + 32 + 4 x 16 = 142 gm.
In 100g of the compound, 44.1 g Na2SO4 contains 55.9 g of H2O.
(∵ 55.9 % is water)
⇒ 1g Na2SO4 contains = (55.9/44.1)g H2O
∴ 142 g Na2SO4 contains = (55.9/44.1) x 142 = 180 g H2O .
Hence, the number of water molecules = 180/18 = 10
(Since the Molecular mass of H2O is 18).
Thus, the value of n is 10 and the formula is Na2SO4.10H2O.
When a metal is burnt, its weight is increased by 24%. The eq. weight of the metal will be:
Equivalent weight of O^2-(oxide) ion is: Molar mass of oxygen/charge on oxygen
= 16/2 = 8 g eq-1
One equivalent of the metal reacts with one equivalent of oxygen to form metal oxide.
Let M be the equivalent weight of the metal.
Therefore, M grams of the metal will react with 8g of oxygen.
The corresponding metal oxide will have a mass of 8 grams more than the mass of metal M. This 8 gram is 24% of M.
i.e. M x 24/100 = 8 gram
Thus, M = 100 x 8/24 = 33.33 g
One mole of potassium chlorate is thermally decomposed and excess of aluminum is burnt in the gaseous product .How many moles of aluminum oxide are formed?
2KClO3 → 2KCl + 3O2
4Al + 3O2 → 2Al2O3
The molar ratio for the first reaction is 2:3 for KClO3:O2
Thus, if 1 mole of KClO3 is decomposed, 1.5 moles of O2 is produced.
The molar ratio for the second reaction is: 3:2 for O2:Al2O3.
3 moles of oxygen produces 2 moles of aluminium oxide.
Since, we have 1.5 moles of oxygen, number of moles of Al2O3 produced =
How much Cl2 at STP is liberated when one mole of KMnO4 reacts with HCl ?
Potassium permanganate reacts with hydrochloric acid according to the following equation:
2KMnO4 + 16HCl → 2KCl + 2MnCl2 + 8H2O + 5Cl2
⇒ 2 moles of KMnO4 produces 5 moles of Cl2
⇒ 1 mole of KMnO4 will produce 2.5 moles of Cl2
∴ Volume of 2.5 moles of Cl2 = 56 litres
Minimum quantity of H2S needed to precipitate 63.5g of Cu2+ is nearly:
Chemical Reaction: Cu2++ H2S → CuS + 2H+
In this reaction, to precipitate 1 mol of Cu2+ we require 1 mol of H2S.
Moles of Cu2+ = 63.5 g / 63.5 g mol-1 = 1 mol
Hence, to precipitate 1 mol of Cu2+ we require 1 mol of H2S
i.e. Mass of H2S = 1 mol x 34 g mol-1 = 34 g
2g of CaCO3 was treated with 0.1M HCl (500 ml). The volume of CO2 evolved at STP after heating the solution is:
10g of CaCO3 on heating gives 5g of the residue (as CaO). The % yield of the reaction is approx:
‘x’ g of KClO3 on decomposition gives ‘y’ ml of O2 at STP. The % purity of KClO3 would be
The balanced eqn. will be:
2KClO3 = 2KCl+3O2
The no. of moles of Oxygen = y/22.4
3 moles of Oxygen is produced from 2 moles of KClO3
⇒ 1 mole will be produced from 2/3 moles of KClO3
∴ y/22.4 will be producing from (2/3) × (y/22.4)
Hence, the mass of KClO3 used = (2/3) × (y/22.4)×M
% Purity = (2 × y × M)/(3 × 22.4 × x)
33.6g of an impure sample of sodium bicarbonate when heated strongly gave 4.4g of CO2. The % purity of NaHCO3 would be:
2NaHCO3 = Na2CO3 + CO2 + H2O
Moles of CO2 = 4.4/44 =0.1
∵ 1 mole of CO2 is produce by 2 Moles of NaHCO3
⇒ Moles of NaHCO3 =0.1×2/1 =0.2
∴ Wt. of NaHCO3 = 0.2 × 84 = 16.8gm
% Purity of NaHCO3 = (16.8/33.6) × 100 = 50%.
0.16g of dibasic acid required 25ml of N/10 NaOH for complete neutralization. Molecular wt. of acid is:
Dibasic acid NaOH;
N1V1 = N2V2
M = 2 × E = 2 × 64 = 128
The molar coefficients of AsO33– and MnO4- in the reaction are:
AsO33– + MnO4Θ → AsO43– + MnO2
Which of the following is a disproportionation reaction?
A disproportionation reaction is a reaction in which a single element undergoes reduction and oxidation
i.e in option d), Cu has changed its oxidation state from +1 to 0 (i.e reduction) and +1 to +2 (i.e oxidation)
Hydrogen and oxygen combine to form H2O2 and H2O containing 5.93% and 11.2 % hydrogen respectively. The data illustrates:
For predicting the law of chemical combination, we have to have find out the ratio of masses in both cases:
Let the total mass in each case be 100 grams.
Mass of hydrogen = 5.93 gm
Mass of oxygen = 100-5.93 = 94.07 gm
The ratio of the mass of hydrogen to the mass of oxygen, H:O = 5.93 : 94.07 = 1 : 16
Mass of hydrogen = 11.2 gm
Mass of oxygen = 100-11.2 = 88.8 gm
The ratio of the mass of hydrogen to the mass of oxygen, H:O = 11.2 : 88.8 = 1 : 32
As the ratio of the mass of hydrogen remains the same in both cases and the ratio of masses of oxygen in both cases (16: 32 or 1: 2) is a whole number ratio.
Therefore, the above data illustrate the law of multiple proportions.