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At 298 K, the vapour pressure of water is 23.75 mm Hg. Calculate the vapour pressure at the same temperature over 5% aqueous solution of urea [CO(NH_(2))_(2)]. |
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Answer» Solution :`5%` aqueous solution of urea means that Mass of solution = 100 g , Mass of SOLUTE, i.e, urea, `w_(2)=5G` `therefore"Mass of solvent, i.e., water "w_(1)=100-5=95 g,` Vapour pressure of pure water `(p^(@))` at 298 K = 23.75 mm Vapour pressure of urea solution =, i.e, `p_(s)=?,` Molar mass of water `(M_(1))="18 g mol"^(-1)` Molar mass of urea `CO(NH_(2))_(2)`, i.e., `M_(2)=12+16+14xx2+2="60 g mol"^(-1)` APPLYING Raoult's law, `(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1))=(w_(2)//M_(2))/(w_(1)//M_(1))=(w_(2))/(M_(2))xx(M_(1))/(w_(1))` Substituting the values, we get `(23.75 - p_(s))/(23.75)=(5)/(60)xx(18)/(95)` `"or"23.75-p_(s)=(5)/(60)xx(18)/(95)xx23.75` `"or"23.75-p_(s)=0.375` `"or"p_(s)=23.375` Thus, the vapour pressure of `5%` urea solution = 23.375 mm |
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