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# Ice crystallises in a hexagonal lattice having volume of the unit cell as 1

Ice crystallises in a hexagonal lattice having volume of the unit cell as 132×10^-24 cm^3.If the density is 0.92 gcm^-3 at a given temperature,then number of H2O molecules per unit cell is

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A)
Density = Mass/Volume

Mass of n molecules of $H_2O$ in hexagonal packing = $\large\frac{18n}{N_{\circ}} = \large\frac{18n}{6.023\times 10^{23}}$

Given Volume = $132\times 10^{-24}cm^3$

Density = $0.92g/cm^3$

Substituting the value we get

$0.92 = \large\frac{18n}{ 6.023\times 10^{23}\times 132\times 10^{-24}}$

$\therefore n = \large\frac{0.92 \times 6.023 \times 10^{23} \times 132\times 10^{-24}}{18}$

$\therefore n = 4$