Home
Class 10
CHEMISTRY
0.1 mol of a basic substance (X) re...

0.1 mol of a basic substance (X) requires `25cm^(3)` of 8.0 `mol//dm^(3)` hydrochloric acid for complete neutralisation .X could be
1. `Al(OH)_(3)`
2. `CuO`
3 . `K_(2)O`
4 . `NaOH`

A

1 and 2

B

2 and 3

C

1 and 4

D

2 and 4

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to determine which of the given basic substances (X) can react with hydrochloric acid (HCl) in the specified mole ratio. Let's break this down step by step. ### Step 1: Calculate the moles of HCl used We are given the molarity and volume of hydrochloric acid: - Molarity of HCl = 8.0 mol/dm³ - Volume of HCl = 25 cm³ = 25 × 0.001 dm³ = 0.025 dm³ Using the formula for moles: \[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 8.0 \, \text{mol/dm}^3 \times 0.025 \, \text{dm}^3 = 0.2 \, \text{mol} \] ### Step 2: Determine the mole ratio of the base to acid We know that 0.1 mol of the basic substance (X) reacts with 0.2 mol of HCl. The mole ratio of base (X) to acid (HCl) is: \[ \text{Mole ratio} = \frac{\text{Moles of X}}{\text{Moles of HCl}} = \frac{0.1}{0.2} = \frac{1}{2} \] ### Step 3: Analyze each option for the mole ratio Now we will check each of the given substances to see if they react with HCl in a 1:2 ratio. 1. **Al(OH)₃ (Aluminum hydroxide)**: \[ \text{Reaction: } \text{Al(OH)}_3 + 3 \text{HCl} \rightarrow \text{AlCl}_3 + 3 \text{H}_2\text{O} \] - Mole ratio: 1 mol Al(OH)₃ reacts with 3 mol HCl (1:3). **Not suitable.** 2. **CuO (Copper(II) oxide)**: \[ \text{Reaction: } \text{CuO} + 2 \text{HCl} \rightarrow \text{CuCl}_2 + \text{H}_2\text{O} \] - Mole ratio: 1 mol CuO reacts with 2 mol HCl (1:2). **Suitable.** 3. **K₂O (Potassium oxide)**: \[ \text{Reaction: } \text{K}_2\text{O} + 2 \text{HCl} \rightarrow 2 \text{KCl} + \text{H}_2\text{O} \] - Mole ratio: 1 mol K₂O reacts with 2 mol HCl (1:2). **Suitable.** 4. **NaOH (Sodium hydroxide)**: \[ \text{Reaction: } \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \] - Mole ratio: 1 mol NaOH reacts with 1 mol HCl (1:1). **Not suitable.** ### Conclusion The substances that can react with HCl in the required 1:2 mole ratio are **CuO** and **K₂O**. ### Final Answer **X could be CuO or K₂O.** ---

To solve the problem, we need to determine which of the given basic substances (X) can react with hydrochloric acid (HCl) in the specified mole ratio. Let's break this down step by step. ### Step 1: Calculate the moles of HCl used We are given the molarity and volume of hydrochloric acid: - Molarity of HCl = 8.0 mol/dm³ - Volume of HCl = 25 cm³ = 25 × 0.001 dm³ = 0.025 dm³ Using the formula for moles: ...
Promotional Banner

Topper's Solved these Questions

  • ACIDS , BASES AND SALTS

    MTG IIT JEE FOUNDATION|Exercise Exercise (Integer /Numberical Value Type)|5 Videos
  • ATOMIC STRUCTURE

    MTG IIT JEE FOUNDATION|Exercise Exercise (Integer/Numerical Value Type)|5 Videos

Similar Questions

Explore conceptually related problems

25cm^(3) of oxalic acid completely neutralised 0.064g of soldium hydroxied. molarity of the oxalic acid solution is

At 25^(@) C for complete combustion of 5 mol propane (C_(3)H_(8)) . The required volume of O_(2) at STP will be .

2.616g of an element (X) is heated with excess of NaNO_(3) and NaOH to produce a basic gas (Y) and Na_(2)XO_(2) . The basic gas liberated exactly requires 5 xx 10^(-3) mol H_(2)SO_(4) for complete neutralisation. If 4 mol of X are completely consumed by caustic soda, find the no of mole of H_(2) gas liberated.

100 ml aqueous solution containing equimolar mixtrue of Ca(OH)_(2) and Al(OH)_(3) requires 0.5 litre of 4M HCl for complete neutralisation. Molarity of Ca(OH)_(2) , in the original solution is :

What is the p^(OH) if the hydrogen ion concentration in solution is 1xx 10^(-3) mol dm^(-3)

The heat of neutralisation of aqueous hydrochloric acid by NaOH si x kcal mol^(-1)of HCI . Calculate the heat of neutralisation per mol of aqueous acetic acid.

1 mol of N_(2) and 2 mol of H_(2) are allowed to react in a 1 dm^(3) vessel. At equilibrium, 0.8 mol of NH_(3) is formed. The concentration of H_(2) in the vessel is

MTG IIT JEE FOUNDATION-ACIDS , BASES AND SALTS -Olympiad /HOTS Corner
  1. What of the following options shows the correct arrangement of diffe...

    Text Solution

    |

  2. Four solutions labelled as P , Q , R and S have pH values 1 , 9,3 and...

    Text Solution

    |

  3. An element which reacts with water to form a solution which turns ph...

    Text Solution

    |

  4. Daivik , a class 10 student studied the reaction between a carbonat...

    Text Solution

    |

  5. The atmosphere of venus is made up of thick white and yellowish clo...

    Text Solution

    |

  6. 0.1 mol of a basic substance (X) requires 25cm^(3) of 8.0 mol//dm...

    Text Solution

    |

  7. Identify the wrong statement .

    Text Solution

    |

  8. The discomform caused by indigestion due to overeasting can be cured b...

    Text Solution

    |

  9. Which of the following graphs shows the change in pH when zinc carbo...

    Text Solution

    |

  10. Plaster of Paris is

    Text Solution

    |

  11. The shining finish to the walls is given by

    Text Solution

    |

  12. Which of the following compounds is alkaline in aqueous medium ?

    Text Solution

    |

  13. This does not possess water of crystallization

    Text Solution

    |

  14. A is an aqueous solution of Acid and B is an aqueous solution of base....

    Text Solution

    |

  15. Which of the following solutions has the lowest pH value ?

    Text Solution

    |

  16. In an experiment , 5 cm^(3) of 1.0 mol//dm^(3) NaOH solution is grad...

    Text Solution

    |

  17. pH of different solutions are given in the table Arrange these...

    Text Solution

    |

  18. If a few drops of a concentrated acid accidentally spill over the hand...

    Text Solution

    |

  19. On passing CO(2) gas in excess in aqueous solution of sodium carbona...

    Text Solution

    |

  20. Which of the following is acidic in nature ?

    Text Solution

    |