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# Chemical Equation MCQ - 1 (Advanced)

## 10 Questions MCQ Test Chemistry for JEE Advanced | Chemical Equation MCQ - 1 (Advanced)

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This mock test of Chemical Equation MCQ - 1 (Advanced) for JEE helps you for every JEE entrance exam. This contains 10 Multiple Choice Questions for JEE Chemical Equation MCQ - 1 (Advanced) (mcq) to study with solutions a complete question bank. The solved questions answers in this Chemical Equation MCQ - 1 (Advanced) quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Chemical Equation MCQ - 1 (Advanced) exercise for a better result in the exam. You can find other Chemical Equation MCQ - 1 (Advanced) extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

### X,Y and Z react in the 1 : 1 : 1 stoichiometric ratio. The concentration of X, Y  and Z where found to vary with time as shown in the figure below Q. Which of the following equilibrium reaction represents the correct variation of concentration with time.

Solution:

Clearly concentration of Y is not changing with the time hence it will be pure solid or liquid. Concentration of X decreasing hence it will be in reactant and Z will be in product.

QUESTION: 2

Solution:

QUESTION: 3

### X,Y and Z react in the 1 : 1 : 1 stoichiometric ratio. The concentration of X, Y  and Z where found to vary with time as shown in the figure below Q. If above equilibrium is established in a 2L container by taking reactants in sufficient amount then how many moles of component Y must have reacted to establish the equilibrium.

Solution:

From the graph we can see that 4 mole/litre of Z is formed therefore      4 × 2 moles of solid will be required.

QUESTION: 4

For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus               pentachloride, etc. the measured densities are found to be less than those calculated from their molecular formula. The observed densities decrease towards a limit as the temperature is raised. This is due to the splitting of the molecules into simpler ones. The process is reversible and is called thermal dissociation.

Examples :

With increase in the number of molecules, the volume increases (pressure  remaining constant) and, in consequence, the density decreases. As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occursthe density reaches its lowest limit.The extent of issociation, ice., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation. Gas density measurements can be used to determine the degree of dissociation. Let us take by general case where one molecule of a substance A splits up into n molecule of B on heating; i.e.,

t = 0 a

t = teq       a  — x               nx

Total no. of moles

Observed molecular weight or molar mass of the mixture

Q.

A sample of mixture of A(g), B(g) and C(g) under equilibrium has a mean molecular weight (observed) is 80.

The equilibrium is

Find the degree of dissociation α for A(g).

Solution:

QUESTION: 5

For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus               pentachloride, etc. the measured densities are found to be less than those calculated from their molecular formula. The observed densities decrease towards a limit as the temperature is raised. This is due to the splitting of the molecules into simpler ones. The process is reversible and is called thermal dissociation.

Examples :

With increase in the number of molecules, the volume increases (pressure  remaining constant) and, in consequence, the density decreases. As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occursthe density reaches its lowest limit.The extent of issociation, ice., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation. Gas density measurements can be used to determine the degree of dissociation. Let us take by general case where one molecule of a substance A splits up into n molecule of B on heating; i.e.,

t = 0 a

t = teq       a  — x               nx

Total no. of moles

Observed molecular weight or molar mass of the mixture

Q.

If the total mass of the mixture in the above case is 300 gm, the moles at C(g) present are.

Solution:

Mols of C = a α

QUESTION: 6

For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus               pentachloride, etc. the measured densities are found to be less than those calculated from their molecular formula. The observed densities decrease towards a limit as the temperature is raised. This is due to the splitting of the molecules into simpler ones. The process is reversible and is called thermal dissociation.

Examples :

With increase in the number of molecules, the volume increases (pressure  remaining constant) and, in consequence, the density decreases. As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occursthe density reaches its lowest limit.The extent of issociation, ice., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation. Gas density measurements can be used to determine the degree of dissociation. Let us take by general case where one molecule of a substance A splits up into n molecule of B on heating; i.e.,

t = 0 a

t = teq       a  — x               nx

Total no. of moles

Observed molecular weight or molar mass of the mixture

Q.

The K for the reaction    is 640 mm at 775 K. The percentage dissociation of N2O4 at equilibrium pressure of 160 mm is :

Solution:

QUESTION: 7

For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus               pentachloride, etc. the measured densities are found to be less than those calculated from their molecular formula. The observed densities decrease towards a limit as the temperature is raised. This is due to the splitting of the molecules into simpler ones. The process is reversible and is called thermal dissociation.

Examples :

With increase in the number of molecules, the volume increases (pressure  remaining constant) and, in consequence, the density decreases. As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occursthe density reaches its lowest limit.The extent of issociation, ice., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation. Gas density measurements can be used to determine the degree of dissociation. Let us take by general case where one molecule of a substance A splits up into n molecule of B on heating; i.e.,

t = 0 a

t = teq       a  — x               nx

Total no. of moles

Observed molecular weight or molar mass of the mixture

Q.

x (degree of dissociation) varies with D/d in the above reaction according to :

Solution:

QUESTION: 8

For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus               pentachloride, etc. the measured densities are found to be less than those calculated from their molecular formula. The observed densities decrease towards a limit as the temperature is raised. This is due to the splitting of the molecules into simpler ones. The process is reversible and is called thermal dissociation.

Examples :

With increase in the number of molecules, the volume increases (pressure  remaining constant) and, in consequence, the density decreases. As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occursthe density reaches its lowest limit.The extent of issociation, ice., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation. Gas density measurements can be used to determine the degree of dissociation. Let us take by general case where one molecule of a substance A splits up into n molecule of B on heating; i.e.,

t = 0 a

t = teq       a  — x               nx

Total no. of moles

Observed molecular weight or molar mass of the mixture

Q.

The equation    is correctly matched for :

Solution:

Hence one mole of reactant should produce total n moles of product.

QUESTION: 9

Match the following

Solution:
QUESTION: 10

Match the following (multiple)

Left column : Represents an equilibrium situratton through a chemical equation and below each equation a stimulus is given which may or may not

disturb the equilibrium situration.

Right coloumn : Represents the responses immediately after the    disturbance is created.

With                                                                      R1 : Rate of forward reaction.

Rb Rate of backward reaction.

Q : Reaction quotient

K : Equilibrium constant

Solution: