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Lowering of vapour pressure in 1 motal aqueous solution at 100^@C is |
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Answer» <P>13.44 mm Hg `m = (x_B XX 1000)/((1-x_B)m_A), 1 = (x_B xx 1000)/((1-x_B) xx 18)` `x_B` = mole fraction of solute, `m_A` = molar mass of solvent `x_B` = 0.0176, `x_A` = 0.9824, p=p `x_A` `=p 760 xx 0.9824=746.624 , Delta p = p-p= 760 -746.624` `~~` 13.4 mm Hg |
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