At 100° C and 1atm , if the density of liquid water is 1.0gcm- 3 and that of water vapour is 0.0006 textgc textm- 3 , then the volume occupied by water molecules in 1 litre of steam at that temperature

Question

At 100° C and 1atm , if the density of liquid water is 1.0gcm- 3 and that of water vapour is 0.0006 textgc textm- 3 , then the volume occupied by water molecules in 1 litre of steam at that temperature (A) 6cm3 (B) 60cm3 (C) 0.6cm3 (D) 0.06cm3

Tardigrade Admin · Accepted Answer

text1L=1000 textmL=1000 textc textm3 textmass= textdensity× textvolume =(0.0006 g cm -3) ×(1000 cm 3)=0.6 g As 18 g of water = text18 c textm3 So 0.6 g of water =0. text6 c textm3 Hence actual volume occupied by molecules =0. text6 c textm3

Q. At $10 0^{\circ} C$ and $1 a t m$ , if the density of liquid water is $1.0 g c m^{- 3}$ and that of water vapour is $0.0006 gc m^{- 3}$ , then the volume occupied by water molecules in 1 litre of steam at that temperature

$6 c m^{3}$

$60 c m^{3}$

$0.6 c m^{3}$

$0.06 c m^{3}$

$1L = 1000 mL = 1000 c m^{3}$
$mass = density \times volume$
$= (0.0006 g c m^{- 3}) \times (1000 c m^{3}) = 0.6 g$
As 18 g of water $= 18 c m^{3}$
So 0.6 g of water $= 0. 6 c m^{3}$
Hence actual volume occupied by molecules $= 0. 6 c m^{3}$

Q. At 100∘C and 1atm , if the density of liquid water is 1.0gcm−3 and that of water vapour is 0.0006gcm−3 , then the volume occupied by water molecules in 1 litre of steam at that temperature

6cm3

60cm3

0.6cm3

0.06cm3