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At `100^(@)C` and 1 atm, if the density of the liquid water is `1.0 g cm^(-3)` and that of water vapour is `0.00006 g cm^(-3)` , then the volume occupied by water molecules in `1 L` steam at this temperature isA. `6 cm^(3)`B. `60 cm^(3)`C. `0.6 cm^(3)`D. `0.06 cm^(3)` |
Answer» Correct Answer - D For water vapours, `P = 0.0006 g c c^(-1)` `:. 0.0006 = ("Mass")/("Volume") = ("Mass")/(1000)` `"Mass" = 1000 xx 0.0006 = 0.6 g` The density of liquid water is `1 g c c^(-1)` So, the volume occupied by water is `("Mass")/("Density"0 = (0.6)/(1) = 0.6 cc` |
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