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oksano4ka [1.4K]
3 years ago
12

Solid cesium iodide has the same kind of crystal structure as CsCl which is pictured below: If the edge length of the unit cell

is 456.2 pm, what is the density of CsI in g/cm3.

Chemistry
1 answer:
Assoli18 [71]3 years ago
3 0

The question is incomplete, here is the complete question:

Solid cesium iodide has the same kind of crystal structure as CsCl which is pictured below: If the edge length of the unit cell is 456.2 pm, what is the density of CsI in g/cm^3

The image is attached below.

<u>Answer:</u> The density of CsI is 9.09g/cm^3

<u>Explanation:</u>

To calculate the density of metal, we use the equation:

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density

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

M = atomic mass of CsI = 259.8 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell = 456.2pm=456.2\times 10^{-10}cm    (Conversion factor:  1cm=10^{10}pm  )

Putting values in above equation, we get:

\rho=\frac{1\times 259.8}{6.022\times 10^{23}\times (456.2\times 10^{-10})^3}\\\\\rho=9.09g/cm^3

Hence, the density of CsI is 9.09g/cm^3

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Relation between degree of dissociation and molar conductivity is as follows.

               \alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}

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Whereas relation between acid dissociation constant and degree of dissociation is as follows.

                     K = \frac{c \times \alpha^{2}}{1 - \alpha}

Putting the values into the above formula we get the following.

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                        = 0.017973 \times 10^{-2}

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Therefore, value of pK_{a} is 3.7454.

                             

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