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A solution containing 15 g of urea (Molar mass =60 g mol^(-1)) per litrea has the same osmotic presure (isotonic) as a solution of glucose (molar mass =180 g mol^(-1)) in water, Calculate themass of glucose present in one litre of the solution. |
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Answer» `"For urea solution "(pi_(1))=((15g)xxRxxT)/((60" g mol"^(-1))XX(1L))` `"For glucose solution"(pi_(2))=(W_(B)RxxT)/((180" g mol"^(-1))xx(1L))` `"For isotinic solutions", pi_(1)=pi_(2)` `therefore((15g))/((60" g mol"^(-1)))=W_(B)/((180" g mol"^(-1)))or BW_(B)=((15g)xx(180" g mol"^(-1)))/((6.0" g mol"^(-1)))=45 g` `therefore "Mass of glucose"(W_(B))=45.0 g.` |
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