Equivalent Mass
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Q. 3 g of an oxide of a metal is converted to form 5 g of its chloride. The equivalent weight of the metal is:
- 33.25 g
- 45.63 g
- 63.57 g
- 98.81 g
Q.
Approximate atomic weight of an element is 26.89. If its equivalent weight is 8.9, the exact atomic weight of element would be
- 26.89
- 26.7
- 8.9
- 17.8
Q.
KMnO4 reacts with ferrous ammonium sulphate according to the equation MnO−4+5Fe2++8H+→Mn2++5Fe3++4H2O, here 10 ml of 0.1 M KMnO4 is equivalent to
- 40 ml of 0.1 M FeSO4
- 50 ml of 0.1 M FeSO4
- 20 ml of 0.1 M FeSO4
- 30 ml of 0.1 M FeSO4
Q. The number of moles of oxalate ions oxidized by one mole of MnO−4 ion in acidic medium.
- 52
- 25
- 35
- 53
Q. For the reaction,
FeS2+KMnO4+H+→Fe3++H2SO4+Mn2++H2O.
If the molar mass of FeS2 is M, then equivalent mass of FeS2 would be equal to:
FeS2+KMnO4+H+→Fe3++H2SO4+Mn2++H2O.
If the molar mass of FeS2 is M, then equivalent mass of FeS2 would be equal to:
- M
- M10
- M11
- M15
Q. 1.6 g of calcium and 2.6 g of zinc when treated with an acid in excess separately produced the same amount of hydrogen gas. Find the equivalent weight of calcium, given zinc has an equivalent weight of 32.6 g.
- 20 g
- 17 g
- 40 g
- 27 g
Q. A solution contains a mixture of Na2CO3 and NaOH. Using phenolphthalein as indicator 35 mL of mixture required 20 mL of 0.80 N HCl for the end point. With methyl orange, 35 mL of solution required 37.5 mL of the same HCl for the end point. What is the weight of Na2CO3 and NaOH respectively in the mixture?
- 1.484 g, 0.08 g
- 1.5 g, 0.5 g
- 0.08 g, 1.484 g
- 0.5 g, 1.5 g
Q. 30 mL of a solution containing 9.15 gm L−1 of an oxalate KxHy(C2O4)z.nH2O are required for titrating 27 mL of 0.12 N NaOH and 36 mL of 0.12 M KMnO4 separately.
The value of x+y+z is:
The value of x+y+z is:
Q.
What is the n-factor / valency factor of H2CO3 ?
1
2
Either a or b
Neither a nor b
Q. 100 mL of N5 NaOH will neutralize
- 0.062 g of H3BO3
- 0.185 g of H3BO3
- 1.24 g of H3BO3
- 0.031 g of H3BO3
Q. Equivalent weight of H3PO4 in each of the reaction will be respectively:
H3PO4+OH−→H2PO−4+H2O
H3PO4+2OH−→HPO2−4+2H2O
H3PO4+3OH−→PO3−4+3H2O
H3PO4+OH−→H2PO−4+H2O
H3PO4+2OH−→HPO2−4+2H2O
H3PO4+3OH−→PO3−4+3H2O
- 98, 49, 32.67
- 49, 98, 32.67
- 98, 32.67, 49
- 32.67, 49, 98
Q. 2.16 g of a metal oxide on strong heating produced 255 mL of oxygen at STP and the residue obtained was a pure metal. Calculate the equivalent weight of the metal.
- 20 g
- 15 g
- 40 g
- 60 g
Q. The mass of 98% pure sulphuric acid required for the neutralization of 2 mol of NaOH is:
(Molar Mass of S = 32 gmol−1, O = 16 gmol−1 and Na = 23 gmol−1)
(Molar Mass of S = 32 gmol−1, O = 16 gmol−1 and Na = 23 gmol−1)
- 100 g
- 98 g
- 200 g
- 196 g
Q. 3.2 g of pyrolusite (MnO2) was treated with 50 mL of 0.5 M oxalic acid and some sulphuric acid. The oxalic acid left undecomposed was raised to 250 mL in a flask. 25 mL of this solution when treated with 0.02 M KMnO4 required 32 mL of the solution. Find the percentage of MnO2 in the sample.
Given: Atomic mass of Mn is 55 g mol−1
Given: Atomic mass of Mn is 55 g mol−1
- 4.5 %
- 11.5 %
- 54.0 %
- 24.4 %
Q. 0.456 g of a metal yielded 0.606 g of its chloride. Calculate the equivalent mass of the metal.
- 91 g
- 61 g
- 31 g
- 108 g
Q. On heating 0.199 g of a metallic oxide in a stream of hydrogen, 0.045 g of water is formed. Find the equivalent weight of the metal.
- 61.2 g
- 31.8 g
- 15.5 g
- 12.7 g
Q. Find the number of moles of KMnO4 needed to oxidise one mole Cu2S in acidic medium. The reaction is KMnO4+Cu2S→Mn2++Cu2++SO2
- 0.4
- 1.6
- 1.2
- 0.5
Q. What is the equivalent mass of HCl in the given reaction?
2KMnO4+16 HCl→2KCl+2MnCl2+5Cl2+8H2O
2KMnO4+16 HCl→2KCl+2MnCl2+5Cl2+8H2O
- 44.2
- 36.5
- 58.4
- none of these
Q.
What is valency factor of acids in a given reaction?
Number of replaceable H+ ions; in the solution
Number of ions of H+ in solution
Number of H+ ions in one molecule of acid
Number of gram - equivalents
Q. Consider a redox reaction:
FeS2+KMnO4+H+→Fe3++SO2−4+Mn2++H2O.
If the molar mass of FeS2 is M, then equivalent mass of FeS2 would be equal to:
FeS2+KMnO4+H+→Fe3++SO2−4+Mn2++H2O.
If the molar mass of FeS2 is M, then equivalent mass of FeS2 would be equal to:
- M
- M5
- M10
- M15
Q. The percentage by mass of chlorine in the chloride of a metal is 80%. 33.375 g of this chloride on vaporization occupies a volume of 5.6 litre at STP. Now select the correct options:
- Valency of the metal is 2
- Molecular weight of the metal chloride is 133.5 u
- Atomic weight of the metal is 27 g
- 9 g of the metal can completely react with 8 g of oxygen
Q.
1.5276 g of CdCl2 was found to contain 0.9367 g of Cd. Atomic weight of Cd is
112.5
122.5
144.5
245
Q. 2.16 g of a metal oxide on strong heating produced 255 mL of oxygen at STP and the residue obtained was a pure metal. Calculate the equivalent weight of the metal.
- 20 g
- 15 g
- 40 g
- 60 g
Q. Potassium acid oxalate K2C2O4.3H2C2O4.4H2O can be oxidized by MnO−4 in acidic medium. Calculate the volume of 0.3 M MnO−4 reacting in acid solution with 10.16 g of the acid oxalate.
(Molar mass of potassium acid oxalate is 508 g mol−1.)
(Molar mass of potassium acid oxalate is 508 g mol−1.)
- 106.7 mL
- 160.0 mL
- 1.06 L
- 0.16 L
Q. Mg can reduce NO−3 to NH3 in basic medium.
NO−3(aq)+Mg(s)+H2O→Mg(OH)2(s)+OH−(aq)+NH3(g)
A 25.0 mL sample of NO−3 solution was treated with Mg(s). The NH3(g) was passed into 100 mL of 0.15 N HCl.
The excess of HCl required 32.10 mL of 0.10 N NaOH for neutralization. What was the molarity of NO−3 ions in the original sample?
NO−3(aq)+Mg(s)+H2O→Mg(OH)2(s)+OH−(aq)+NH3(g)
A 25.0 mL sample of NO−3 solution was treated with Mg(s). The NH3(g) was passed into 100 mL of 0.15 N HCl.
The excess of HCl required 32.10 mL of 0.10 N NaOH for neutralization. What was the molarity of NO−3 ions in the original sample?
- 3.025
- 0.5005
- 0.47125
- 0.3025
Q. Calculate the mass of K2Cr2O7 required to produce 254 g I2 from KI solution.
K2Cr2O7+KI→Cr3++I2
(Molar mass of K2Cr2O7 is 294 g mol−1)
K2Cr2O7+KI→Cr3++I2
(Molar mass of K2Cr2O7 is 294 g mol−1)
- 49 g
- 98 g
- 9.8 g
- 49 g
Q. 1.20 g sample of Na2CO3 and K2CO3 was dissolved in water to form 100 mL of a solution. 20 mL of this solution required 40 mL of 0.1 N HCl for complete neutralization. Calculate the weight of precipitate in the mixture if another 20 mL of this solution is treated with excess of BaCl2.
- 0.394 g
- 0.115 g
- 0.596 g
- 0.604 g
Q. Find the equivalent weight of KMnO4 in the given reaction, if the molar mass of MnO−4 ion is 'M' ?
MnO−4→Mn4+ (neutral medium)
MnO−4→Mn4+ (neutral medium)
- 22.7
- 12.7
- 32.7
- 52.7
Q. Find the equivalent weight of thiosulphate ion from the following reaction:
2S2O2−3(Reducing agent)→S4O2−6+2e−
2S2O2−3(Reducing agent)→S4O2−6+2e−
- M
- 2M
- M2
- 3M
Q. 3.2 g of pyrolusite (MnO2) was treated with 50 mL of 0.5 M oxalic acid and some sulphuric acid. The oxalic acid left undecomposed was raised to 250 mL in a flask. 25 mL of this solution when treated with 0.02 M KMnO4 required 32 mL of the solution. Find the percentage of MnO2 in the sample.
Given: Atomic mass of Mn is 55 g mol−1
Given: Atomic mass of Mn is 55 g mol−1
- 4.5 %
- 24.4 %
- 54.0 %
- 11.5 %