English

How Many Grams of Magnesium Will Have the Same Number of Atoms as 6 Grams of Carbon ? (Mg = 24 U ; C = 12 U ) - Science

Advertisements
Advertisements

Question

How many grams of magnesium will have the same number of atoms as 6 grams of carbon ?
(Mg = 24 u ; C = 12 u )

Short/Brief Note
Advertisements

Solution

Mass of carbon = 6g

We know that equal moles of all substances contain equal number of atoms (1 mole of all substances contains 6.022 × 1023 atoms).

The first step will be to convert 6g  carbon into moles.

1 mole of carbon = 12g

12g  carbon = 1 mole

Therefore, 6g carbon = 0.5 moles

Since, equal moles of all substances contain equal number of atoms, 0.5 moles of carbon will have the same number of atoms as present in 0.5 moles of magnesium.

The next step will be to find out the mass of 0.5 moles of magnesium in grams.

1 mole of magnesium = 24g

0.5 moles of magnesium = 24 × 0.5 = 12g

Hence, 12g magnesium will contain the same number of atoms as present in 6g  carbon.

shaalaa.com
  Is there an error in this question or solution?
Chapter 3: Atoms and Molecules - Very Short Answers [Page 174]

APPEARS IN

Lakhmir Singh Chemistry [English] Class 9
Chapter 3 Atoms and Molecules
Very Short Answers | Q 49 | Page 174

RELATED QUESTIONS

If the number of electrons in an ion Z3– is 10, the atomic number of element Z will be :


The atomic number of an element E is 16. The number of electrons in its ion E2– will be :


The atom of an element X contains 17 protons, 17 electrons and 18 neutrons whereas the atom of an element Y contains 11 protons, 11 electrons and 12 neutrons.

  1. What type of ion will be formed by an atom of element X ? Write the symbol of ion formed.
  2. What will be the number of (i) protons (ii) electrons, and (iii) neutrons, in the ion formed from X ?
  3. What type of ion will be formed by an atom of element Y ? Write the symbol of ion formed.
  4. What will be the number of (i) protons (ii) electrons, and (iii) neutrons, in the ion formed from Y ?
  5. What is the atomic mass of (i) X, and (ii) Y ?
  6. What could the elements X and Y be ?

What is the numerical value of Avogadro number ?


What name is given to the number 6.022 × 1023 ?


What name is given to this number ?


If 12 gram of carbon has x atoms, then the number of atoms in 12 grams of magnesium will be :


Which of the following has the maximum number of atoms ?


What weight of oxygen gas will contain the same number of molecules as 56 g of nitrogen gas ?

(O = 16 u ; N = 14 u)


What mass of nitrogen, N2, will contain the same number of molecules as 1.8 g of water, H2O ?

(Atomic masses : N = 14 u ; H = 1 u ; O = 16 u)


The mass of one atom of an element X is 2.0 × 10–23 g.

Calculate the atomic mass of element X.

What could element X be ?


Which has more number of atoms?

100g of N2 or 100 g of NH3


A gold sample contains 90% of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold?


Compute the difference in masses of one mole each of aluminium atoms and one mole of its ions. (Mass of an electron is 9.1×10–28 g). Which one is heavier?


A silver ornament of mass ‘m’ gram is polished with gold equivalent to 1% of the mass of silver. Compute the ratio of the number of atoms of gold and silver in the ornament.


. A sample of ethane (C2H6) gas has the same mass as 1.5 ×1020 molecules of methane (CH4). How many C2H6 molecules does the sample of gas contain?


In the atom of an element X, 6 electrons are present in the outermost shell. If it acquires noble gas configuration by accepting requisite number of electrons, then what would be the charge on the ion so formed?


Match the names of the Scientists given in column A with their contributions towards the understanding of the atomic structure as given in column B

  (A)   (B)
(a) Ernest Rutherford (i) Indivisibility of atoms
(b) J.J.Thomson (ii) Stationary orbits
(c) Dalton (iii) Concept of nucleus
(d) Neils Bohr (iv) Discovery of electrons
(e) James Chadwick (v) Atomic number
(f) E. Goldstein (vi) Neutron
(g) Mosley (vii) Canal rays

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×