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Ice crystallizes in hexagonal lattice. At the low temperature at which structure was determined the lattice constants were $a = 4.53A^{\large\circ}$ and C = $a = 7.41A^{\large\circ}$ .Calculate the no. of $H_2O$ molecules present in a unit cell. (Density of ice = 0.92g/cm)

$(a)\;2\qquad(b)\;3\qquad(c)\;4\qquad(d)\;5$

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Volume of unit cell V = Area of $base\times height$
$=2\times\large\frac{\sqrt3}{4}a^2\times c$
$\large\frac{\sqrt3}{2}\times(4.53\times10^{-8})^2\times7.4\times10^{-8}$
$=13.17\times10^{-23}cm^3$
Mass of unit cell = $V\times\rho$
$=13.17\times10^{-23}\times0.92$
$=1.21\times10^{-22}g$
Let there are n effective water molecules in the given unit cell. Then n times the mass of each water molecule would be the mass of unit cell.
$\therefore 1.21\times10^{-22} = n\times\large\frac{18}{6.023\times10^{23}}$
$\therefore n =4$
Hence answer is (c)
answered Feb 24, 2014 by sharmaaparna1
 

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