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Lowering of vapour pressure due to a solute in `1` molal aqueous solution at `100^(@)C` is a.`13.44 mm Hg` ,b. `14.14 mm Hg` ,c.`13.2 mm Hg` ,d. `35.2 mm Hg`A. 13.44 mm HgB. 14.12 mm HgC. 31.2 mm HgD. 35.2 mm Hg |
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Answer» Correct Answer - A `p = p_(A) + p_(B)` `p = p_(A)^(@) x_(A) + p_(B)^(@) x_(B)` `= 120 xx (2)/(4) + 80 xx (2)/(4)` `= 60 + 40 = 100` torr `y_(B) =` mole fraction of B in the vapour phase `= (p_(B))/(p_("total")) = (40)/(100) = (2)/(5)` |
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