How many atoms of niobium are present in 2.43 ~g if it forms bcc structure with density 9 ~g ~cm ⁻³ and volume of unit cell 2.7 10⁻²³ ~cm ^3 ?

Step-by-step solution

Let number of atoms are N Mass of one atom =2.43 / N d = Z mass of one atom Volume of unit cell [ Z =2 for BCC ] 9= 2 2.43 / N 2.7 10⁻²³ ~N = 2 2.43 2.7 10⁻²³ 9 =2 10²²