Question

a sample of a pure element that has a mass of 1.00 g contains 4.39×10^21 atom. what is the element. ..​

Answer 1

Hope this helps u co-relate with the given ans

Answer 2

Element is Barium

Explanation:

• Given No of atoms = 4.39 ×$10^{21}$
• mass of element = 1g
• one mole of element contains avogadro's number of atoms i.e  6.022×$10^{23}$ and the mass of it will be its atomic mass
• $\frac{GIven mass}{Atomic mass}$ = $\frac{Given no.of atoms}{Avogadro's number}$
• Atomic mass = $Given mass \frac{Avogadro's  number}{Given no.of atoms}$
• Atomic mass = 1×$\frac{6.022}{4.39}$×100
• Solving we get Atomic mass = 137g
• This atomic mass corresponds to element Barium