Ice crystallises in a hexagonal lattice. At the low temperature, the lattice constants were a = 4.53 mathringA and b = 7.41 mathringA. How many H2O molecules are contained in a unit cell? [d(ice) = 0.92 g / cm]

Question

Ice crystallises in a hexagonal lattice. At the low temperature, the lattice constants were a = 4.53 mathringA and b = 7.41 mathringA. How many H2O molecules are contained in a unit cell? [d(ice) = 0.92 g / cm] (A) 4 (B) 3 (C) 2 (D) 1

Tardigrade Admin · Accepted Answer

Volume of the unit cell, V= area of rhombus base × b =(a2 sin 60°) × b =(4.53)2 × (√3/2) × 7.41 × 10-24 cm 3 =132 × 10-24 cm 3 Thus, the mass of unit cell =V × d=132 × 10-24 × 0.92 × 6.02 × 1023 =73 g This is close to 4 times the molecular weight of H 2 O . So, there are four molecules of H 2 O per unit.

Q. Ice crystallises in a hexagonal lattice. At the low temperature, the lattice constants were $a = 4.53 \overset{˚}{A}$ and $b = 7.41 \overset{˚}{A}$ . How many $H_{2} O$ molecules are contained in a unit cell? $[d (i ce) = 0.92 g / c m]$

$4$

$3$

$2$

$1$

Q. Ice crystallises in a hexagonal lattice. At the low temperature, the lattice constants were a=4.53A˚ and b=7.41A˚. How many H2​O molecules are contained in a unit cell? [d(ice)=0.92g/cm]

4

3

2

1

Volume of the unit cell, V= area of rhombus base ×b =(a2sin60∘)×b =(4.53)2×23​​×7.41×10−24cm3 =132×10−24cm3 Thus, the mass of unit cell =V×d=132×10−24×0.92×6.02×1023 =73g This is close to 4 times the molecular weight of H2​O. So, there are four molecules of H2​O per unit.

Q. Ice crystallises in a hexagonal lattice. At the low temperature, the lattice constants were $a = 4.53 \overset{˚}{A}$ and $b = 7.41 \overset{˚}{A}$ . How many $H_{2} O$ molecules are contained in a unit cell? $[d (i ce) = 0.92 g / c m]$

$4$

$3$

$2$

$1$