Volume occupied by one molecule of water (density = 1 g cm -3 ) is

Question

Volume occupied by one molecule of water (density = 1 g cm -3 ) is (A) 5.5 × 10-23 cm 3 (B) 9.0 × 10-23 cm 3 (C) 6.023 × 10-23 cm 3 (D) 3.0 × 10-23 cm 3

Tardigrade Admin · Accepted Answer

As water is liquid its density =1 g / mL i.e., 1 g of H 2 O have volume =1 mL Mass of one molecule =(18/6.023 × 1023) g ∴ (18/6.023 × 1023) g of H 2 O have volume =(18/6.022 × 1023) mL =3.0 × 10-23 mL

Q. Volume occupied by one molecule of water (density = $1 g c m^{- 3}$ ) is

$5.5 \times 1 0^{- 23} c m^{3}$

$9.0 \times 1 0^{- 23} c m^{3}$

$6.023 \times 1 0^{- 23} c m^{3}$

$3.0 \times 1 0^{- 23} c m^{3}$

As water is liquid its density $= 1 g / m L$
i.e., $1 g$ of $H_{2} O$ have volume $= 1 m L$
Mass of one molecule $= \frac{18}{6.023 \times 1 0 ^{23}} g$
$∴ \frac{18}{6.023 \times 1 0 ^{23}} g$ of $H_{2} O$ have volume
$= \frac{18}{6.022 \times 1 0 ^{23}} m L = 3.0 \times 1 0^{- 23} m L$

Q. Volume occupied by one molecule of water (density = 1gcm−3 ) is

5.5×10−23cm3

9.0×10−23cm3

6.023×10−23cm3

3.0×10−23cm3