Stoichiometry MCQ - 1 (Advance) - JEE MCQ

Chloride of an element is given by the formula MCI, and it is 100% ionised in 0.01 M aqueous solution.Then

If 100 ml of 1M H₂SO₄ solution is mixed with 100 ml of 98%(wlw) H₂SO₄ solution (d = 0.1 gmlml) then :

3 moles of the gas C₂H₆is mixed with 60 gm of this gas and 2.4 x 10²⁴ molecules of the gas is removed. The left over gas is combusted in the presence of excess oxygen then : (NA = 6 x 10²³) (Density of water = 1 gm/ml)

Which of the following contains the same number of molecules?

For producing 4 gram moles of HNO3, how many grams of NO2 are required?

Choose the correct statement : 

632 g of sodium thiosulphate (Na2S₂O₃) reacts with copper sulphate to form cupric thiosulphate which is reduced by sodium thiosulphate to give cuprous compound which is dissolved in excess of sodium thiosulphate to form a complex compound sodium cuprothiosulphate ( Na₄[Cu₆(S₂O₃)₅] ).

         Sodium cuprothiosulphaté

In this process, 0.2 mole of sodium cuprothiosulphate is formed. (O = 16, Na = 23, S = 32)

The average oxidation states of sulphur in Na₂S₂O₃ and Na₂S₄O₆ are respectively.

632 g of sodium thiosulphate (Na2S₂O₃) reacts with copper sulphate to form cupric thiosulphate which is reduced by sodium thiosulphate to give cuprous compound which is dissolved in excess of sodium thiosulphate to form a complex compound sodium cuprothiosulphate ( Na₄[Cu₆(S₂O₃)₅] ).

         Sodium cuprothiosulphaté

In this process, 0.2 mole of sodium cuprothiosulphate is formed. (O = 16, Na = 23, S = 32)

Q.

Moles of sodium thiosulphate reacted and unreacted after the reaction are respecticvely.

632 g of sodium thiosulphate (Na2S₂O₃) reacts with copper sulphate to form cupric thiosulphate which is reduced by sodium thiosulphate to give cuprous compound which is dissolved in excess of sodium thiosulphate to form a complex compound sodium cuprothiosulphate ( Na₄[Cu₆(S₂O₃)₅] ).

         Sodium cuprothiosulphaté

In this process, 0.2 mole of sodium cuprothiosulphate is formed. (O = 16, Na = 23, S = 32)

Q.

If instead of given amount of sodium thiosulphate, 2 moles of sodium thiosulphate along with 3 moles of CuSO₄were taken initially. Then moles of sodium cuprothiosulphate formed is

We know that balancing of a chemical equation is entirely based on law of conservation of mass. However the concept of Principle of Atom Conservation (POAC) can also be related to law of conservation of mass in a chemical reaction. So, POAC can also act as a technique for balancing a chemical equation. For example, for a reaction

2AB3 → AB + C2

0n applying POAC for, B & C and relating the 3 equations, we get : ( nx :  number of 
moles of X) Thus, the coefficients of ABC₃ , AB & C₂ in the balanced chemical equation will be 2,2 & 3 respectively and the balanced chemical equation can be represented as

ABC₃ → AB + 3C₂

_{Now answer the following questions :}

_Q.

Which of the following relation is correct regarding the numerical coefficients p, q , r in the balanced chemical equation :

pA + qB2 → rA2B5

We know that balancing of a chemical equation is entirely based on law of conservation of mass. However the concept of Principle of Atom Conservation (POAC) can also be related to law of conservation of mass in a chemical reaction. So, POAC can also act as a technique for balancing a chemical equation. For example, for a reaction

2AB3 → AB + C2

0n applying POAC for, B & C and relating the 3 equations, we get : ( nx :  number of 
moles of X) Thus, the coefficients of ABC₃ , AB & C₂ in the balanced chemical equation will be 2,2 & 3 respectively and the balanced chemical equation can be represented as

ABC₃ → AB + 3C₂

_{Now answer the following questions :}

Q.

If the weight ratio of C and 0₂ present is 1 : 2 and both of reactants completely consume and form C0 and C0₂ and we will obtain a gasous mixture of C0 and CO₂. What would be the weight ratio of C0 and C0₂ in mixture.

We know that balancing of a chemical equation is entirely based on law of conservation of mass. However the concept of Principle of Atom Conservation (POAC) can also be related to law of conservation of mass in a chemical reaction. So, POAC can also act as a technique for balancing a chemical equation. For example, for a reaction

ABC₃ → AB + C₂

0n applying POAC for, B & C and relating the 3 equations, we get : ( nx :  number of 
moles of X) Thus, the coefficients of ABC₃ , AB & C₂ in the balanced chemical equation will be 2,2 & 3 respectively and the balanced chemical equation can be represented as

2ABC₃ → 2AB + 3C₂

_{Now answer the following questions :}

Q.

If the atomic masses of X and Y are 10 & 30 respectivetly, then the mass of XY3 formed when 120g of Y2 reacts completely with X is : Reaction X + Y2 ¾¾® XY3

Oleum is considered as a solution of S0₃ in H₂S0₄ , which is obtained by passing 803 in solution of H₂SO₄. When 100 g sample of oleum is diluted with desired weight of H₂0 then the total mass of H₂SO₄ obtained after dilution is known as % labelling in oleum.
For example, a oleum bottle labelled as '109% H₂SO₄ means the 109g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of Hp which combines with all the free 803 to form H₂SO₄ as 

S0₃ + H₂0 → + H₂SO₄.

Q.

What is the % of free S0₃ in an oleum that is labelled as '104.5% H₂SO₄’?

Oleum is considered as a solution of S0₃ in H₂S0₄ , which is obtained by passing 803 in solution of H₂SO₄. When 100 g sample of oleum is diluted with desired weight of H₂0 then the total mass of H₂SO₄ obtained after dilution is known as % labelling in oleum.
For example, a oleum bottle labelled as '109% H₂SO₄ means the 109g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of Hp which combines with all the free 803 to form H₂SO₄ as 

S0₃ + H₂0 → + H₂SO₄.

Q.

If excess water is added into a 100 g bottle sample labelled as '' 112% H₂SO₄ '' and is reacted with 5.3 g Na₂CO₃, then find the volume of C0₂ evolved at 1 atm pressure and 300 k temperature after the completion of the reaction : [R = 0.0821 L atom mol^–1 K^–1]

H₂SO₄ + Na₂CO₃ → Na₂SO₄ + H₂0 + CO₂

Oleum is considered as a solution of S0₃ in H₂S0₄ , which is obtained by passing 803 in solution of H₂SO₄. When 100 g sample of oleum is diluted with desired weight of H₂0 then the total mass of H₂SO₄ obtained after dilution is known as % labelling in oleum.
For example, a oleum bottle labelled as '109% H₂SO₄ means the 109g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of Hp which combines with all the free 803 to form H₂SO₄ as 

S0₃ + H₂0 → + H₂SO₄.

Q.

1 g of oleum sample is diluted with water.The solution required 54 ml o f 0.4 N NaOH for  compete neutralization.The % of free S03 in the sample is :

Volumetric analysis is based on the principle of equivalence a that minloces lead in the ratio of their equivalents. At the equivalence point of the reaction, involving the reactahts A and B : Number of gram equivalents of A = Number of gram equivalents of B..
If VA ml of solution A having normality NA react just completely with VB ml  of solution B having normalityNB then

Equation(1), called normality equation is very useful in numerical of volumetric analysis. Equivalent masses of different substances :

Basically the equivalent mass of a substance is defined as the pads by seam

ctituhich combine with or displace 1.0078 parts (≈1part) by mass of hydrogen, 8 parts by mansetaisiosaid 35.5 parts by mass of chlorine.

                Mass of a substance expressed in gram equal to its equivalent man is' gam equivalent mass. Equivalent mass of a substance is not constant but depends upon the maw i malice the substance participates.

                        Equivalent mass of an acid in acid-base reaction is its mass in grannehidicodoinisooleaf replaceable Fi ions (= 1.0078 g≈1g).0n the other hand, equivalent mass of a bagels istolonellfacontairis 1 mole of replaceable 0H^– ions. 1 g equivalent mass each of an acid and base an ionclinegins saint and 1 mole of water ( = 18 g).

Equivalent mass of an oxidising agent is its mass which gains 1 mole at ellecismitcan be obtained by dividing the molecular mass or formula mass by the total decrease in aididditaiiilberef one or more elements per molecule.

                0n the other hand, equivalent mass of a reducing agent is the mass of the substance which loses 1 mole of electorns. It can be calculated by dividing the molecular or formula  mass of the subtance by the total increase in oxidation number of one or more elements per molecular or formula mass.

Q.

When 5.00 g of a metal is strongly heated, 9.44 g of its oxide is obtained. The equivalent mass of the metal is

Volumetric analysis is based on the principle of equivalence a that minloces lead in the ratio of their equivalents. At the equivalence point of the reaction, involving the reactahts A and B : Number of gram equivalents of A = Number of gram equivalents of B..
If VA ml of solution A having normality NA react just completely with VB ml  of solution B having normalityNB then

Equation(1), called normality equation is very useful in numerical of volumetric analysis. Equivalent masses of different substances :

Basically the equivalent mass of a substance is defined as the pads by seam ctituhich combine with or displace 1.0078 parts (≈1part) by mass of hydrogen, 8 parts by mansetaisiosaid 35.5 parts by mass of chlorine.

Mass of a substance expressed in gram equal to its equivalent man is' gam equivalent mass. Equivalent mass of a substance is not constant but depends upon the maw i malice the substance participates.

Equivalent mass of an acid in acid-base reaction is its mass in grannehidicodoinisooleaf replaceable Fi ions (= 1.0078 g≈1g).0n the other hand, equivalent mass of a bagels istolonellfacontairis 1 mole of replaceable 0H^– ions. 1 g equivalent mass each of an acid and base an ionclinegins saint and 1 mole of water= 18 g). Equivalent mass of an oxidising agent is its mass which gains 1 mole at ellecismitcan be obtained by dividing the molecular mass or formula mass by the total decrease in aididditaiiilberef one or more elements per molecule.

0n the other hand, equivalent mass of a reducing agent is the mass of the substance which loses 1 mole of electorns. It can be calculated by dividing the molecular or formula  mass of the subtance by the total increase in oxidation number of one or more elements per molecular or formula mass.

Q.

How many gram of phosphoric acid (H₃PO₄) would be required to neutralize 58 g of magnesium hydroxide?

Volumetric analysis is based on the principle of equivalence a that minloces lead in the ratio of their equivalents. At the equivalence point of the reaction, involving the reactahts A and B : Number of gram equivalents of A = Number of gram equivalents of B..
If VA ml of solution A having normality NA react just completely with VB ml  of solution B having normalityNB then

Equation(1), called normality equation is very useful in numerical of volumetric analysis. Equivalent masses of different substances :

Basically the equivalent mass of a substance is defined as the pads by seam ctituhich combine with or displace 1.0078 parts (≈1part) by mass of hydrogen, 8 parts by mansetaisiosaid 35.5 parts by mass of chlorine.

Mass of a substance expressed in gram equal to its equivalent man is' gam equivalent mass. Equivalent mass of a substance is not constant but depends upon the maw i malice the substance participates.

Equivalent mass of an acid in acid-base reaction is its mass in grannehidicodoinisooleaf replaceable Fi ions (= 1.0078 g≈1g).0n the other hand, equivalent mass of a bagels istolonellfacontairis 1 mole of replaceable 0H^– ions. 1 g equivalent mass each of an acid and base an ionclinegins saint and 1 mole of water= 18 g). Equivalent mass of an oxidising agent is its mass which gains 1 mole at ellecismitcan be obtained by dividing the molecular mass or formula mass by the total decrease in aididditaiiilberef one or more elements per molecule.

0n the other hand, equivalent mass of a reducing agent is the mass of the substance which loses 1 mole of electorns. It can be calculated by dividing the molecular or formula  mass of the subtance by the total increase in oxidation number of one or more elements per molecular or formula mass.

Q.

Which of the following solutions, when mixed with 100 ml of 0.05 M NaOH, will give a neutral solution?

Volumetric analysis is based on the principle of equivalence a that minloces lead in the ratio of their equivalents. At the equivalence point of the reaction, involving the reactahts A and B : Number of gram equivalents of A = Number of gram equivalents of B..
If VA ml of solution A having normality NA react just completely with VB ml  of solution B having normalityNB then

Equation(1), called normality equation is very useful in numerical of volumetric analysis. Equivalent masses of different substances :

Basically the equivalent mass of a substance is defined as the pads by seam ctituhich combine with or displace 1.0078 parts (≈1part) by mass of hydrogen, 8 parts by mansetaisiosaid 35.5 parts by mass of chlorine.

Mass of a substance expressed in gram equal to its equivalent man is' gam equivalent mass. Equivalent mass of a substance is not constant but depends upon the maw i malice the substance participates.

Equivalent mass of an acid in acid-base reaction is its mass in grannehidicodoinisooleaf replaceable Fi ions (= 1.0078 g≈1g).0n the other hand, equivalent mass of a bagels istolonellfacontairis 1 mole of replaceable 0H^– ions. 1 g equivalent mass each of an acid and base an ionclinegins saint and 1 mole of water= 18 g). Equivalent mass of an oxidising agent is its mass which gains 1 mole at ellecismitcan be obtained by dividing the molecular mass or formula mass by the total decrease in aididditaiiilberef one or more elements per molecule.

0n the other hand, equivalent mass of a reducing agent is the mass of the substance which loses 1 mole of electorns. It can be calculated by dividing the molecular or formula  mass of the subtance by the total increase in oxidation number of one or more elements per molecular or formula mass.

Q.

100 ml of 0.1 M H₃ P0₂ is titrated with 0.1 M NaOH. The volume of NaOH needed will be

Volumetric analysis is based on the principle of equivalence a that minloces lead in the ratio of their equivalents. At the equivalence point of the reaction, involving the reactahts A and B : Number of gram equivalents of A = Number of gram equivalents of B..
If VA ml of solution A having normality NA react just completely with VB ml  of solution B having normalityNB then

Equation(1), called normality equation is very useful in numerical of volumetric analysis. Equivalent masses of different substances :

Basically the equivalent mass of a substance is defined as the pads by seam ctituhich combine with or displace 1.0078 parts (≈1part) by mass of hydrogen, 8 parts by mansetaisiosaid 35.5 parts by mass of chlorine.

Mass of a substance expressed in gram equal to its equivalent man is' gam equivalent mass. Equivalent mass of a substance is not constant but depends upon the maw i malice the substance participates.

Equivalent mass of an acid in acid-base reaction is its mass in grannehidicodoinisooleaf replaceable Fi ions (= 1.0078 g≈1g).0n the other hand, equivalent mass of a bagels istolonellfacontairis 1 mole of replaceable 0H^– ions. 1 g equivalent mass each of an acid and base an ionclinegins saint and 1 mole of water= 18 g). Equivalent mass of an oxidising agent is its mass which gains 1 mole at ellecismitcan be obtained by dividing the molecular mass or formula mass by the total decrease in aididditaiiilberef one or more elements per molecule.

0n the other hand, equivalent mass of a reducing agent is the mass of the substance which loses 1 mole of electorns. It can be calculated by dividing the molecular or formula  mass of the subtance by the total increase in oxidation number of one or more elements per molecular or formula mass.

Q.

9.8 g of an acid on treatment with excess of an alkali forms salt along with 3.6 g of water. What is the equivalent mass of the acid?

Molality : It is defined as the moles of the solute present in 1kg of the solvent. It is denoted by 'm'.

Let W_A grams of the solute of molecular mass mA be present in W_B grams of the solvent, then 

Relation between mole fraction and Molality : 

Q.

If the ratio of the mole fraction of a solute is changed from -3 to -2 in the 800 g of solvent then the ratio ofmolality will be :

Molality : It is defined as the moles of the solute present in 1kg of the solvent. It is denoted by 'm'.

Let W_A grams of the solute of molecular mass mA be present in W_B grams of the solvent, then 

Relation between mole fraction and Molality : 

Q.

The mole fraction of the solute in the 12 molal solution of Na₂CO₃ is :

Molality : It is defined as the moles of the solute present in 1kg of the solvent. It is denoted by 'm'.

Let W_A grams of the solute of molecular mass mA be present in W_B grams of the solvent, then 

Relation between mole fraction and Molality : 

Q.

What is the quantity of water that should be added to 16 gm, methanol to make the mole fraction of methanol as 0.25 –

Molality : It is defined as the moles of the solute present in 1kg of the solvent. It is denoted by 'm'.

Let W_A grams of the solute of molecular mass mA be present in W_B grams of the solvent, then 

Relation between mole fraction and Molality : 

Q.

A 300 gm, 30% (wlw) NaOH solution is mixed with 500 gm 40% (wow) NaOH solution. What is % (wlw) NaOH. if density of final solution is 2 gm/ml.

Molality : It is defined as the moles of the solute present in 1kg of the solvent. It is denoted by 'm'.

Let W_A grams of the solute of molecular mass mA be present in W_B grams of the solvent, then 

Relation between mole fraction and Molality : 

Q.

What is the molality of final solution obtained in the above problem

Match the reacting mixture in coloumn-I with the reagent in coloumn – Il

Column – I                        Coloumn – Il