A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm^(-3)? The molar mass of the metal is (N_(A)" Avogadro's constant" = 6.02times10^(23) mol^(-1))

A metal has a fcc lattice. The edge length of the unit cell is 404 pm.

1.	A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm^(-3)? The molar mass of the metal is (N_(A)" Avogadro's constant" = 6.02times10^(23) mol^(-1))
Answer» <html><body><p>`40" g "<a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>^(-1)`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/30-304807" style="font-weight:bold;" target="_blank" title="Click to know more about 30">30</a>" g "mol^(-1)`<br/>`27" g "mol^(-1)`<br/>`20" g "mol^(-1)`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`"Density " rho = (ZtimesM)/(a^(3)timesN_(A)times10^(-30))` <br/> `therefore 2.72 = (4timesM)/((404^(3))times(6.02times10^(<a href="https://interviewquestions.tuteehub.com/tag/23-294845" style="font-weight:bold;" target="_blank" title="Click to know more about 23">23</a>))times10^(-30))`<br/> `therefore M = (2.72times(404)^(3)times(6.02times10^(23))times10^(-30))/4`<br/> =`27 g"" mol^(-1)`</body></html>