Question

The number of iron ($Fe$) atoms present in a pure sample of solid iron with a mass of ten grams is equal to- .
A.$(10.0) \times (55.9) \times (6.02 \times {10^{23}})$ atoms
B.$\dfrac{{6.02 \times {{10}^{23}}}}{{(10.0)(55.9)}}$
C.$\dfrac{{(10.0)(6.02 \times {{10}^{23}})}}{{(55.9)}}$
D.$\dfrac{{55.9}}{{(10.0)(6.02 \times {{10}^{23}})}}$
E.\[\dfrac{{10.0}}{{(55.9)(6.02 \times {{10}^{23}})}}\]

Answer 1

Hint: Atomic number and mass number both are primary qualities of an iota of the component. In the event that you know the atomic and mass number of an iota of the component, at that point you can without much of a stretch figure the quantity of neutrons present in the core of the particle. By knowing atomic number and mass number of a component we can anticipate numerous properties of the component.

Complete step by step answer:
 What is Avogadro’s number?
 The amount of units in a solitary mole of any substance is called Avogadro's number or Avogadro's steady. It is identical to $6.022140857 \times {10^{23}}$. The units may be electrons, particles, particles, or iotas, dependent upon the character of the reaction and the possibility of the substance.
 What is the atomic mass of iron and what are its significance?
 It is our vulnerability of the real outright isotopic bounties of iron that limit its prescribed atomic mass of $55.845u$ to just five huge digits. The atomic masses of different isotopes can be resolved to an exceptionally serious extent of precision utilizing mass spectrometry. Iron, with four normally happening isotopes, is no exception. The masses of the other iron isotopes are known to a similar serious extent of exactness. Iron is a definitive final result of heavenly atomic combination and the isotopic bounties of iron differ from test to test far and wide relying upon the earthly source from which the example is taken (e.g., unadulterated iron metal versus mineral stores or organic examples, geological cause of the example, and etc..)
The quantity of iron ( $Fe$) particles present in a pure sample of solid iron with a mass of 10 grams is equivalent to $\dfrac{{(10.0)(6.02 \times {{10}^{23}})}}{{(55.9)}}$.

Thus, the alternative (C) is the right answer.


Note: $\dfrac{{10.0}}{{55.9}}$ speaks to the proportion of mass of iron to molar mass. It speaks to the number of moles of iron. At the point when the number of moles of iron is duplicated with Avogadro’s number $6.021 \times {10^{23}}$, we get a number of iron molecules .