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
So ,number of moles in 2.24L
Total moles is 0.2 moles.
Total molecules in 0.2moles
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 gm 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
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*2g of iodine requires 63*10g of HNO3
Thus, 5g of HNO requires:
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.
To know the mass of Na2SO4 containing 55.9% H2O , we subtract 55.9 from 100, 100 - 55.9 = 44.1
Therefore 44.1 g Na2SO4 contains = 55.9 gm H2O
So, 1gm Na2SO4 contains = 55.9 / 44.1 gm H2O
142 gm Na2SO4 contains = 55.9 /44.1 x 142 = 180 gm H2O .
Number of water molecules = 180 / 18 (Since Molecular mass of H2O is 18).
= 10 water molecules .
Hence 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:
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 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
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 if produced from 2moles of KClO3 ...then 1 mole will produce from 2/3moles of KClO3 ...And y/22.4 will be producing from 2/3×y/22.4
,..no. of moles of KClO3 = 2/3×y/22.4
So..mass of KClO3 used =2/3×y/22.4×M(mol. wt.)
%purity = 2/3×y/22.4×M divide the whole to 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
Then moles of NaHCO3 =0.1×2/1 =0.2
And wt. Of NaHCO3 = 0.2 × 84 = 16.8gm
%purity of NaHCO3 = 16.8 × 100/33.6 =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),in Cu2O, Cu has an oxidation state of +1 and in Cu it has 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 the cases:
For that, 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
Ratio of mass of hydrogen to mass of oxygen,
i.e. H:O = 5.93 : 94.07
H:O = 1 : 16
Mass of hydrogen = 11.2 gm
Mass of oxygen = 100-11.2 = 88.8 gm
Ratio of mass of hydrogen to mass of oxygen,
i.e. H:O = 11.2 : 88.8
H:O = 1 : 32
As the ratio of mass of hydrogen remains same in both the cases and the ratio of masses of oxygen in both the cases (16 : 32 or 1 : 2) is a whole number ratio.
Therefore, the above data illustrates the law of multiple proportions.
Hence, option (4) is correct.