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At $100^{\large\circ}C$ and 1 atm , if the density of liquid water is 1.0g $cm^{-3}$ and that of water vapour is $0.0006gcm^{-3}$ , then the volume occupied by water molecules in 1 litre of stream at that temperature is :

$(a)\;6cm^3\qquad(b)\;60cm^3\qquad(c)\;0.6cm^3\qquad(d)\;0.06cm^3$

1 Answer

Given
Density of liquid water = 1.0g $cm^{-3}$
Density of water vapour = 0.0006g $cm^{-3}$
Mass of 1000mL steam = $1000\times0.0006 = 0.6g$
Since, Volume of liquid water = $\large\frac{\text{Mass of 1000mL steam}}{\text{Density of liquid water}} = \large\frac{0.6g}{1g\;cm^{-3}}$
$\therefore$ Volume of liquid water = $0.6 cm^3$
Hence answer is (c)
answered Feb 13, 2014 by sharmaaparna1
edited Apr 2, 2014 by mosymeow_1
 

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