Chemistry, 09.07.2019 17:00 jaallen3679

You discover a new alloy made up of cu and ni atoms. it has a bcc structure, where the ni atoms are located at the corners and the cu atoms are located in the center of each cell. the radius of the cu atoms is 0.13 nm and the radius of the ni atom is 0.15 nm. calculate the density of this structure. assume that cu has an atomic weight of 40 g/mol and ni has an atomic weight of 60 g/mol.

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Answer from: ravenl1001

Answer : The Density of the structure = 10.23005 g

Solution : Given,

Atomic weight of Cu = 40 g/mol

Atomic weight of Ni = 60 g/mol

Radius of Cu atom = 0.13 nm

Radius of Ni atom = 0.15 nm

Type of crystal structure = BCC

The number of atom in the BCC unit cell (Z) = 2

Avogadro's number ( $N_{A}$ ) = 6.022 × $10^{23}$ $mol^{-}$

First we have to find the Edge length of unit cell i.e (a)

According to the question, in the BCC structure Ni atoms are present at the corners. So, we will use the radius of Ni atom for the calculation of edge length of unit cell. Image given below.

Edge length of unit cell (a) = 2 × radius of Ni atom = 2 × 0.15 nm = 0.30 nm = 3 × $10^{-8}$ cm

Converstion, 1 nm = $10^{-7}$ cm

Volume of unit cell = $a^{3}$ = $(3\times 10^{-8})^{3}$ = 27 × $10^{-24}$ $cm^{3}$

In an alloy of Cu & Ni , Cu atom is present at centre and eight Ni atom at the eight corners of the cube.

Thus, there is $\frac{1}{8} \times 8$ = 1atom of Ni and 1 atom of Cu

Total Molar mass of unit cell of Cu-Ni alloy (M) = molar mass of Cu + molar mass of Ni

= (40 + 60) g/mol

= 100 g/mol

Formula used :

$\text{Density} = \frac{Z\times M}{N_{A}\times a^{3}}$

Put all the values in above formula, we get

$\text {Density}= \frac{2\times 100}{(6.022\times 10^{23})\times( 27 \times 10^{-24})}$

= 10.23005 g

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