Question

a crystalline solid AB adopts sodium chloride type structure with edge length of the unit cell as 745pmand formula mass of 74.5amu. the density of the crstallinecompound is
a)2.16g/cm^3 b)0.99g/cm^3 c)1.88g/cm^3 d)1.197g/cm^3
please tell the value of 1amu.

Answer 1

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Answer 2

Answer : The correct option is, (d) $1.197g/Cm^{3}$

Solution : Given,

Number of atom in unit cell of FCC (Z) = 4

Edge length = $745pm=745\times 10^{-10}cm$     $(1pm=10^{-10}cm)$

Atomic mass of NaCl (M) = 74.5 amu = 74.5 g/mole    (1 amu = 1 g/mole)

Avogadro's number $(N_{A})=6.022\times 10^{23} mol^{-1}$

Formula used :  

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$      .............(1)

where,

$\rho$ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number  

a = edge length of unit cell

Now put all the values in above formula (1), we get

$\rho=\frac{4\times (74.5g/mol)}{(6.022\times 10^{23}mol^{-1}) \times(745\times 10^{-10}Cm)^3}=1.197g/Cm^{3}$

Therefore, the density of the unit cell is, $1.197g/Cm^{3}$