Following equilibrium is established at temperature T . A ( g ) ⇌ B ( g ) + C ( g ) at equilibrium 1 M 2 M 2 M If the volume of the vessel is doubled then find the equilibrium concentration of each species. (Given that : 40 = 6 . 324 )

Given A ( g ) ⇌ B ( g ) + C ( g ) at equilibrium 1 M 2 M 2 M K C = B C A We know Concentration Molarity = Number of moles Volume of solution When the volume of solution is doubled then the equilibrium concentration of each species gets halved, as the number of moles of species remains the same. Concentration of each species after volume is doubled. A ' = n A 2 V = A 2 = 1 2 = 0 . 5 M B ' = n B 2 V = B 2 = 2 2 = 1 M C ' = n C 2 V = C 2 = 2 2 = 1 M

Following equilibrium is established at temperature T . A ( g ) ⇌ B ( g ) + C ( g ) at equilibrium 1 M 2 M 2 M If the volume of the vessel is doubled then find the equilibrium concentration of each species. (Given that : 40 = 6 . 324 )

Given A ( g ) ⇌ B ( g ) + C ( g ) at equilibrium 1 M 2 M 2 M K C = B C A We know Concentration Molarity = Number of moles Volume of solution When the volume of solution is doubled then the equilibrium concentration of each species gets halved, as the number of moles of species remains the same. Concentration of each species after volume is doubled. A ' = n A 2 V = A 2 = 1 2 = 0 . 5 M B ' = n B 2 V = B 2 = 2 2 = 1 M C ' = n C 2 V = C 2 = 2 2 = 1 M

A dilute solution of H 2 SO 4 is made by adding 5 mL of 3 N H 2 SO 4 to 245 mL of water. Find the normality and molarity of the diluted solution.

On dilution, N i V i = N f V f N f = 3 × 5 245 = 15 245 = 0 . 06 N N = M × n -factor n -factor of H 2 SO 4 = 2 M = N 2 = 0 . 06 2 = 0 . 03 M

20 ml of H 2 O 2 , after acidification with dilute H 2 SO 4 , required 30 ml of N 12 KMnO 4 for complete oxidation. Determine the strength of H 2 O 2 solution.

N 1 V 1 = N 2 V 2 Here N 1 = Strength of KMnO 4 V 1 = Volume of KMnO 4 N 2 = Strength of H 2 O 2 V 2 = Volume of H 2 O 2 Since complete oxidation is taking place, N 1 V 1 = N 2 V 2 . N 2 = N 1 V 1 V 2 = 1 × 30 12 × 20 = 1 8 Now strength = N 2 × equivalent mass = 1 8 × 17 = 2 . 12 g / L

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Following equilibrium is established at temperature $T$ .

$\begin{matrix} A (g) & ⇌ & B (g) & + & C (g) \\ at equilibrium & 1 M & 2 M & 2 M \end{matrix}$

If the volume of the vessel is doubled then find the equilibrium concentration of each species. (Given that : $\sqrt{40} = 6.324$ )

Important Questions on Some Basic Concepts of Chemistry

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Alpha Question Bank for Engineering: Chemistry>Chapter 1 - Some Basic Concepts of Chemistry>Exercise-1>Q 80

Molar concentration of

96 g

O_{2}

contained in a

2

litre vessel is -

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Alpha Question Bank for Engineering: Chemistry>Chapter 1 - Some Basic Concepts of Chemistry>Exercise-1>Q 81

The reaction

A (g) + B (g) ⇌ C (g) + D (g)

is studied in a one litre vessel at

250 ° C

. The initial concentration of

A

was

3 n

and that of

B

was

n

. When equilibrium was attained, equilibrium concentration of

C

was found to the equal to the equilibrium concentration of

B

. What is the concentration of

D

at equilibrium?

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Alpha Question Bank for Engineering: Chemistry>Chapter 1 - Some Basic Concepts of Chemistry>Exercise-1>Q 82

The equilibrium

N_{2} (g) + O_{2} (g) ⇌ 2 NO (g)

estabilished in a reaction vessel of

2.5 L

capacity. The amounts of

N_{2}

and

O_{2}

taken at the start were respectively

2

moles and

4

moles. Half a mole of nitrogen has been used up at equilibrium. The molar concentration of nitric oxide is :

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Alpha Question Bank for Engineering: Chemistry>Chapter 1 - Some Basic Concepts of Chemistry>Exercise-1>Q 83

A dilute solution of

H_{2} {SO}_{4}

is made by adding

5 mL

3 N

H_{2} {SO}_{4}

245 mL

of water. Find the normality and molarity of the diluted solution.

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Alpha Question Bank for Engineering: Chemistry>Chapter 1 - Some Basic Concepts of Chemistry>Exercise-1>Q 84

20 ml

H_{2} O_{2}

, after acidification with dilute

H_{2} {SO}_{4}

, required

30 ml

\frac{N}{12} {KMnO}_{4}

for complete oxidation. Determine the strength of

H_{2} O_{2}

solution.

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Alpha Question Bank for Engineering: Chemistry>Chapter 1 - Some Basic Concepts of Chemistry>Exercise-1>Q 85

x g

of the metal gave

y

g

of its oxide. Hence equivalent weight of the metal

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Alpha Question Bank for Engineering: Chemistry>Chapter 1 - Some Basic Concepts of Chemistry>Exercise-1>Q 86

Equivalent wt. of $H_{3} {PO}_{4}$ in each of the reaction will be respectively -

$H_{3} {PO}_{4} + {OH}^{-} \to H_{2} {PO}_{4^{-}}^{} + H_{2} O$
$H_{3} {PO}_{4} + 2 {OH}^{-} \to {HPO}_{4}^{2 -} + 2 H_{2} O$
$H_{3} {PO}_{4} + 3 {OH}^{-} \to {PO}_{4}^{3 -} + 3 H_{2} O$

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Alpha Question Bank for Engineering: Chemistry>Chapter 1 - Some Basic Concepts of Chemistry>Exercise-1>Q 87

How many millilitres of

0.1 N H_{2} {SO}_{4}

solution will be required for complete reaction with a solution containing

0.125 g

of pure

{Na}_{2} {CO}_{3}

Following equilibrium is established at temperature T. A(g)⇌B(g)+C(g)at equilibrium1M2M2M If the volume of the vessel is doubled then find the equilibrium concentration of each species. (Given that : 40=6.324)

Important Questions on Some Basic Concepts of Chemistry

Learn and Prepare for any exam you want !

Following equilibrium is established at temperature $T$ .

$\begin{matrix} A (g) & ⇌ & B (g) & + & C (g) \\ at equilibrium & 1 M & 2 M & 2 M \end{matrix}$

If the volume of the vessel is doubled then find the equilibrium concentration of each species. (Given that : $\sqrt{40} = 6.324$ )