Solveeit Logo

Question

Question: Which among the following has the largest number of atoms? [A] 0.5g atom of Cu [B] 0.635g of Cu ...

Which among the following has the largest number of atoms?
[A] 0.5g atom of Cu
[B] 0.635g of Cu
[C] 0.25 mol of Cu atom
[D] 1g of Cu

Explanation

Solution

Hint: To find the number of atoms in any case remember that-
The molar mass (g) of any molecule contains NA{{N}_{A}}number of atoms of that molecule. Also, each mole contains NA{{N}_{A}}number of atoms. So, the number of atoms in gm or in moles can be calculated easily.

Complete step by step solution:
We know, the molar mass of Cu is 63.5g
I.e. 63.5g of copper contains Avogadro’s number (NA{{N}_{A}}) atoms of copper
\therefore 63.5g copper= 6.023×10236.023\times {{10}^{23}}atoms of copper.
So, following unitary method, we get-
0.635g of copper=(6.023×1023×0.635)÷63.5(6.023\times {{10}^{23}}\times 0.635)\div 63.5atoms of copper= 6.023×10216.023\times {{10}^{21}}atoms of copper
Similarly, for 1g of copper we have,
1g of copper= (6.023×1023×1)÷63.5(6.023\times {{10}^{23}}\times 1)\div 63.5=9.485×10219.485\times {{10}^{21}}atoms of copper
For 0.5g atom of Cu,
Number of atoms= 0.5×6.022×10230.5\times 6.022\times {{10}^{23}}= 3.011×10233.011\times {{10}^{23}}atoms of copper
Now, for 0.25 moles of copper,
\because 1 mole of copper= 6.023×10236.023\times {{10}^{23}}atoms of copper
\therefore 0.25 moles of copper= 0.25×6.023×10230.25\times 6.023\times {{10}^{23}}=1.50575×10231.50575\times {{10}^{23}}atoms of copper.
From the above calculations it is clear that 0.5g atom of Copper has the highest number of atoms present.
Therefore, option [A] 0.5g atom of Cu atom is the correct answer.

ADDITIONAL INFORMATION:
According to Avogadro’s Law, 1 mole of every gas occupies 22.4 litres of volume at STP (0{{0}^{\circ }}C temperature and 1atm pressure) and contains Avogadro’s number6.023×10236.023\times {{10}^{23}} of particles.

NOTE: It is important here to remember the value of the Avogadro’s number and also the molar mass of copper, the rest can be calculated easily. Also note that in option [A] the unit it g atom
g atom= 1 mole = NA{{N}_{A}}atoms.
Whereas, in options [B] and [D] the unit is in grams so we have to divide it by the molar mass. And again option [C] is in mole units so it will be simply multiplied to the Avogadro’s number.