Home
Class 11
CHEMISTRY
A sample of peanut oil weighing 1.5763 g...

A sample of peanut oil weighing `1.5763 g` is added to `25 mL "of" 0.4210 M KOH`. After saponification is complete `8.46 mL "of" 0.2732 M H_(2)SO_(4)` is needed to neutralize excess `KOH`. The saponification number of peanut oil is:

A

`209.6`

B

`108.9`

C

`98.9`

D

`218.9`

Text Solution

Verified by Experts

The correct Answer is:
A
Meq.of `KOH` added `=25xx0.4210=10.525`
Meq. of `KOH` left `=8.46xx0.2732xx2=4.623`
`:.` Meq. of `KOH` used by oil`=10.525-4.623=5.902`
or `(w)/(56)xx1000=5.902`
or `w_(KOH)=0.3305g`
`:.` Saponification no. `=wt.of KOH` used in `mg per g` of oil
`=(0.3305)/(1.5763)xx1000=209.6`
Promotional Banner

Topper's Solved these Questions

  • MOLE AND EQUIVALENT CONCEPT

    P BAHADUR|Exercise Exercise(3B)Objective problems|15 Videos

  • MOLE AND EQUIVALENT CONCEPT

    P BAHADUR|Exercise Exercise (4) Objective problems|20 Videos

  • MOLE AND EQUIVALENT CONCEPT

    P BAHADUR|Exercise Exercise (2) prevous year numberical problems|25 Videos

  • IONIC EQUILIBRIUM

    P BAHADUR|Exercise Exercise|85 Videos

  • RADIO ACTIVITY

    P BAHADUR|Exercise Exercies 9|99 Videos

Similar Questions

Explore conceptually related problems

A sample of peanut oil weighing 2g is added to 25mL of 0.40M KOH . After saponification is complete, 8.5mL of 0.28M H_2SO_4 is needed to nuetralize excess of KOH . The saponification number of peanut oil is : (saponification number is defined as the milligrams of KOH consumed by 1g of oil)

10ml of 1 M H_(2)SO_(4) will completely neutralise

……………. ml of 0.1 M H_(2)SO_(4) is required to neutralize 50 ml of 0.2 M NaOH Solution :

________ ml of 0.1 M H_(2)SO_(4) is required to neutralise 50 ml of 0.2 M NaOH solution

What volume of 0.108 M H_(2)SO_(4) is required to neutralize 25.0 mL of 0.145 M KOH?

P BAHADUR-MOLE AND EQUIVALENT CONCEPT-Exercise (3A) Objective problems:
  1. A solution contains Na(2)CO(3) and NaHCO(3). 10 mL of the solution req...

    Text Solution

    |

  2. 0.7 g "of" Na(2)CO(3).xH(2)O were dissolved in water and the volume wa...

    Text Solution

    |

  3. A sample of peanut oil weighing 1.5763 g is added to 25 mL "of" 0.4210...

    Text Solution

    |

  4. Eq.wt. of an acid salt NaHSO(4) is:

    Text Solution

    |

  5. When a metal is burnt, its weight is increased by 24%. The equivalent ...

    Text Solution

    |

  6. 0.71 g of chlorine combines with certain weight of a metal giving 1.11...

    Text Solution

    |

  7. How many grams of phosphoric acid would be needed to neutralise 100 g ...

    Text Solution

    |

  8. 100 mL of mixture of NaOH and Na(2)SO(4) is neutralised by 10 mL of 0....

    Text Solution

    |

  9. 0.05 moles of NaHCO(3) will react with how many equivalent of Mg(OH)(2...

    Text Solution

    |

  10. 0.078 g Al(OH)(3). is dehydrated to Al(2)O(3). The Al(2)O(3) so obtain...

    Text Solution

    |

  11. Assuming 100% ionisation, the solution having highest normality is:

    Text Solution

    |

  12. 100 mL of a sample of hard water requires 25.1 mL of 0.02 N H(2)SO(4) ...

    Text Solution

    |

  13. The equivalent weight of potash alum (K(2)SO(4).Al(2)(SO(4))(3).24H(...

    Text Solution

    |

  14. Vapour density of a metal chloride is 6.6. Its oxide contains 53% meta...

    Text Solution

    |

  15. 100 mL of 0.1 M solution of H(2)SO(4) is used to prepare 0.05N solutio...

    Text Solution

    |

  16. Which does not change on dilution?

    Text Solution

    |

  17. 20 mL of 0.1M H(3)BO(3) solution on complete netralisation requires …....

    Text Solution

    |

  18. Volume strength of H(2)O(2) labelled is 10 vol. What is normality of H...

    Text Solution

    |

  19. Find the weight of H(2) SO(4) in 1200 mL of a solution of 0.2 N streng...

    Text Solution

    |

  20. Calculate the volume of 0.5 M H2SO4 required to dissolve 0.5 g of copp...

    Text Solution

    |