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Standard XII

Chemistry

Density

An ionic soli...

Question

An ionic solid $F e O$ crystallizes into a cubic structure and has edge length equal to 5.0 A. If the density of crystal is 4 g/cm $^{3}$ , then the relation between formula mass $(X)$ and mass of a unit cell $(Y)$ is:

X = Y

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X = 4 Y

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Y = 2 X

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Y = 4 X

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Solution

The correct option is D $Y = 4 X$
Density $= ρ = \frac{Z \times M_{A}}{N_{A} \times a^{3}}$
Mass of $F e O = 56 + 16 = 72$
Given,
$Z = ?$
$a = 5.0 ˚ A$
$M_{A} = 72$
$ρ = 4$ $g / c m^{3}$
$N_{A} = 6.023 \times 10^{2} 3$
$4 = \frac{Z \times 72}{6.023 \times 125 \times 10^{- 24}}$
$Z = 4$
Therefore, appoximately $4$ molecules of $F e O$ are present in a unit cell.
$Y = 4 X$

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An ionic solid FeO crystallizes into a cubic structure and has edge length equal to 5.0 A. If the density of crystal is 4 g/cm3, then the relation between formula mass (X) and mass of a unit cell (Y) is:

The correct option is D Y=4XDensity =ρ=Z×MANA×a3Mass of FeO=56+16=72Given,Z=?a=5.0˚AMA=72ρ=4 g/cm3NA=6.023×10234=Z×726.023×125×10−24Z=4Therefore, appoximately 4 molecules of FeO are present in a unit cell.Y=4X

An ionic solid $F e O$ crystallizes into a cubic structure and has edge length equal to 5.0 A. If the density of crystal is 4 g/cm $^{3}$ , then the relation between formula mass $(X)$ and mass of a unit cell $(Y)$ is: