Home
Class 12
CHEMISTRY
The simplest formula of a compound conta...

The simplest formula of a compound containing 50% of an element `X` (atomic weight 10) and 50% of element `Y` (atomic weight 20) is:

A

`XY_(2)`

B

`X_(2)Y`

C

`X_(2)Y_(3)`

D

`XY_(3)`

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • SOME BASIC CONCEPTS OF CHEMISTRY

    VMC MODULES ENGLISH|Exercise IN - CHAPTER EXERCISE - D|28 Videos
  • SOME BASIC CONCEPTS OF CHEMISTRY

    VMC MODULES ENGLISH|Exercise IN - CHAPTER EXERCISE - E|28 Videos
  • SOME BASIC CONCEPTS OF CHEMISTRY

    VMC MODULES ENGLISH|Exercise IN - CHAPTER EXERCISE - B|9 Videos
  • SOME BASIC CONCEPT OF CHEMISTRY

    VMC MODULES ENGLISH|Exercise IMPECCABLE|49 Videos
  • STATES OF MATTER

    VMC MODULES ENGLISH|Exercise IMPECCABLE|50 Videos

Similar Questions

Explore conceptually related problems

Objective question (only one correct). i. The simplest formula of a compound containing 50% of element A (atomic mass 10) and 50% of element B (atomic mass 20) is: a. AB b. A_(2) B c. A_(2) B_(3) d. AB_(3)

What is the simplest formula of acompound that contains 0.25 g atom of sillicon per 0.50 g atom of oxygen.

Determine the empirical formula of a compound containing 47.9% potassium, 5.5% beryllium and 46.6% fluorine by mass. (Atomic weight of Be=9, F = 19, K = 39).

One atom of an element x weight 6.643xx10^(-23)g . Number of moles of atom in 20 kg is :

One atom of an element x weight 6.643xx10^(-232)g . Number of moles of atom in 20 kg is :

A compound contains 20% sulphur. The molecular weight of the compound could be

The Mw of a oxide of an element is 44. The Ew of the element is 14. The atomic weight of the element is

The relative abundance of two isotopes of an element with atomic weight 85 and 87 is 75% and 25% respectively. Then calculate the average atomic weight of element

The molecular weight of chloride of an element is 79. The Ew of the elements is 3.82. The atomic weight of the elements is

An element forms two oxides containing 50% and 40% of the element by mass. Prove that the results are in agreement with the law of multiple proportions.

VMC MODULES ENGLISH-SOME BASIC CONCEPTS OF CHEMISTRY-IN - CHAPTER EXERCISE - C
  1. A gaseous alkane is exploded with O2. The volume of O2 required for ...

    Text Solution

    |

  2. A polystryrene, having formula Br(3) C(6) H(2) (C(8) H(8))(n), was per...

    Text Solution

    |

  3. The simplest formula of a compound containing 50% of an element X (ato...

    Text Solution

    |

  4. How many molecules of water are present as water of crystallisation in...

    Text Solution

    |

  5. The volume in litres of CO(2) liberated at STP when 10 grams of 90% pu...

    Text Solution

    |

  6. In the reaction 4A + 2B + 3C to A4B2C3, the number of moles of product...

    Text Solution

    |

  7. The mass of sodium hydroxide produced when "175.5 g of NaCl" reacts wi...

    Text Solution

    |

  8. 30g Mg and 30g O(2) are reacted and the residual mixture contains:

    Text Solution

    |

  9. Calculate the mass of lime (CaO) obtained by heating 200kg of 95% pure...

    Text Solution

    |

  10. A silver coin weighing 11.34 g was dissolved in nitric acid When sodiu...

    Text Solution

    |

  11. Consider the following reaction sequence: S(8)(s) + 8O(2)(g) to 8SO(...

    Text Solution

    |

  12. 4Fe+3O(2) rarr 2Fe(2)O(3) (reference to iron )

    Text Solution

    |

  13. 30 ml of a mixture of oxygen (O(2)) and ozone (O(3)) was heated till o...

    Text Solution

    |

  14. Polythene can be produced from calcium carbide according to the follow...

    Text Solution

    |

  15. The density of liquid ethanol is 0.7892 g/ml at 20^@ C. If 1.2 mol of ...

    Text Solution

    |

  16. 200 ml of an aqueous solution of glucose (C(6)H(12)O(6)) has molarity ...

    Text Solution

    |

  17. The normality of HCl solution with a density of 1.19gm/ml containing 3...

    Text Solution

    |

  18. 10L of hard water required 5.6g of lime for removing hardness. Hence t...

    Text Solution

    |

  19. Permanent hardness is due to CI^(ɵ) and SO4^(2-) of Mg^(2+) and Ca^(2+...

    Text Solution

    |

  20. Equal moles of water and user are taken in a flask. What is mass perce...

    Text Solution

    |