Ice crystallises in a hexagonal lattice having volume of the unit cell as 122 × 10- 24 cm3 . If its density is 0.90 g cm- 3 at a given temperature, then number of H2O molecules per unit cell is/are

Question

Ice crystallises in a hexagonal lattice having volume of the unit cell as 122 × 10- 24 cm3 . If its density is 0.90 g cm- 3 at a given temperature, then number of H2O molecules per unit cell is/are (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

(m a s s/v o l u m e)= d Mass of n molecules of H2O in hexagonal packing=(18 n/N0)=(18 n/6.02 × 1023) = (18 n/6.02 × 1 023 × 122 × 1 0- 22)= 0.90 n=4

Q. Ice crystallises in a hexagonal lattice having volume of the unit cell as $122 \times 1 0^{- 24} c m^{3}$ . If its density is $0.90 g c m^{- 3}$ at a given temperature, then number of $H_{2} O$ molecules per unit cell is/are

$1$

$2$

$3$

$4$

$\frac{ma ss}{v o l u m e} = d$
Mass of $n$ molecules of $H_{2} O$ in $h e x a g o na l p a c kin g = \frac{18 n}{N _{0}} = \frac{18 n}{6.02 \times 1 0 ^{23}}$
$= \frac{18 n}{6.02 \times 1 0 ^{23} \times 122 \times 1 0 ^{- 22}} = 0.90$
$n = 4$

Q. Ice crystallises in a hexagonal lattice having volume of the unit cell as 122×10−24cm3 . If its density is 0.90gcm−3 at a given temperature, then number of H2​O molecules per unit cell is/are

1

2

3

4

volumemass​=d Mass of n molecules of H2​O in hexagonalpacking=N0​18n​=6.02×102318n​ =6.02×1023×122×10−2218n​=0.90 n=4

Q. Ice crystallises in a hexagonal lattice having volume of the unit cell as $122 \times 1 0^{- 24} c m^{3}$ . If its density is $0.90 g c m^{- 3}$ at a given temperature, then number of $H_{2} O$ molecules per unit cell is/are

$1$

$2$

$3$

$4$

$\frac{ma ss}{v o l u m e} = d$
Mass of $n$ molecules of $H_{2} O$ in $h e x a g o na l p a c kin g = \frac{18 n}{N _{0}} = \frac{18 n}{6.02 \times 1 0 ^{23}}$
$= \frac{18 n}{6.02 \times 1 0 ^{23} \times 122 \times 1 0 ^{- 22}} = 0.90$
$n = 4$