Home
Class 11
CHEMISTRY
The mass of 60% HCI required for the net...

The mass of 60% HCI required for the netralisation of 10L of `0.1 MKOH` is

A

`60.8 g `

B

`21.9 g`

C

`100 g`

D

`219 g`

Text Solution

Verified by Experts

The correct Answer is:
A
M.eq. of `KOH = 10000 xx 0.1 = 1000`
`therefore m . eq. of HCl therefore W _(BCl) = (1000)/(1000 ) xx 36.5 = 36.5 g` pure
60 g pure sample `-= 100 g` sample HCl `therefore 36.5` g pure sample `= (100)/(60) xx 36.5 g`
Sample HCl `= 60.8 g`
Promotional Banner

Topper's Solved these Questions

  • SOME BASIC CONCAPTS OF CHEMISTRY

    BRILLIANT PUBLICATION|Exercise LEVEL-II|50 Videos

  • SOME BASIC CONCAPTS OF CHEMISTRY

    BRILLIANT PUBLICATION|Exercise LEVEL-III (Single Correct Answer Type)|14 Videos

  • SOME BASIC CONCAPTS OF CHEMISTRY

    BRILLIANT PUBLICATION|Exercise LEVEL-III (Linked Comprehension Type)|6 Videos

  • S-BLOCK ELEMENTS

    BRILLIANT PUBLICATION|Exercise LEVEL-III (Linked Comprehension Type)|9 Videos

  • SOME BASIC CONCEPTS OF CHEMISTRY

    BRILLIANT PUBLICATION|Exercise LEVEL-I|55 Videos

Similar Questions

Explore conceptually related problems

Calculate the mass of CaO required to remove the hardness of 10^(6) litre of water containing 1.62 g of calcium bicarbonate per litre.

calculate the mass of NaOH required to make 500mL of 0.5 M aqueous solution. (molecular mass of NaOH = 40).

Magnesium reduced NO_(3) according to the equation given below: NO_(3)^(-) + Mg + 3H_(2)O rarr Mg(OH)_(2)+ OH + NH_(3) In basic medium, 25 ml sample of NO_(2) was reacted with Mg. The ammonia gas evolved was passed through 50 ml of 1.15 N HCI. The excess HCI required 32.1 ml of 0.1 M NaOH for neutralization. The molarity of NO_(3) in the original sample is

Calculate the mass of NaOH required to make 500 mL of 0.5 molar aqueous solution (Molecular mass of NaOH=40)

Calculate the mass of NaOH required to make 500 ml of 0.5 M aqueous solution. (molecular mass of NaOH = 40)

Calculate the mass of calcium oxide required that react with 852 g of P_(4)O_(10) .

Gold number of Gum Arabic is 0.15. The amount of Gum Arabic required to protect 100 mL of red gold sol from coagulation by 10 mL of 10% NaCl solution is

BRILLIANT PUBLICATION-SOME BASIC CONCAPTS OF CHEMISTRY -LEVEL-I
  1. 8.6 g of an element when treated with excess of acid produce 1900ml of...

    Text Solution

    |

  2. Molarity of 560 g pure H(2)O at 4^(@)C is

    Text Solution

    |

  3. The mass of 60% HCI required for the netralisation of 10L of 0.1 MKOH ...

    Text Solution

    |

  4. Cortisone is a molecular substance containing 21 atoms of carbon per m...

    Text Solution

    |

  5. A sample of ammonium phosphate (NH(3))(4) PO(4) contains 3.18 moles of...

    Text Solution

    |

  6. Consider the following laws of chemical combination with examples: I...

    Text Solution

    |

  7. 1 g" urea "(NH(2)CO NH(2)), 1g acetic acid (CH(3) COOH) and 1g of form...

    Text Solution

    |

  8. The molarity of a solution obtained by mixing 750mL of 0.5 M HCl with ...

    Text Solution

    |

  9. 40% W"/"V NaCl solution (specific gravity =1.12) is equivalent to : ...

    Text Solution

    |

  10. In compound A, 1.0 g nitrogen combines with 0.57 g of Oxygen. In compo...

    Text Solution

    |

  11. Which one of the phrases would be incorrect to use?

    Text Solution

    |

  12. Which one of the best example of law of conservation of mass?

    Text Solution

    |

  13. Number of atoms present in 4.25g of 'NH3' is

    Text Solution

    |

  14. 3g of a hydrocarbon on combustion in excess of oxygen produces 8.8 g" ...

    Text Solution

    |

  15. A metal oxide is reduced by heating it in a stream of hydrogen. It is ...

    Text Solution

    |

  16. The volume occupied by one molecule of water (density = 1 gcm^(-3))

    Text Solution

    |

  17. Common salt obtained from sea-water contains 95% NaCl by mass. The app...

    Text Solution

    |

  18. The volume of H2 evolved at STP when 0.9 g of Al(molar mass : 27 g mol...

    Text Solution

    |

  19. The mass of NaBrO3 required to prepare 150 mL of 0.75 N of a solution ...

    Text Solution

    |

  20. The expression converting mole fraction of a solute into molarity of s...

    Text Solution

    |