Question

5% solution of glucose is isotonic with a solution of urea m=50 hence the weight of urea present in the solution is​

Answer 1

i)5% solution of glucose is isotonic with a solution of urea m=50 hence the weight of urea present in the solution is1.368

ii) Let us take the weight of urea present in the solution be x and the molecular weight of urea when calculated is found to be 342.

iii) As it is an isotonic solution so using the following method,

π1=π2  

(or) (n1/V1)ST=(n2/V2)ST

(or) n1/V1=n2/V2

∴C1=C2

⇒w1/m1V1=w2/m2V2

w1/m1V1[sugar]=w2/m2V2[urea]

∴5/(342×100/1000)=x /(50×100/1000

)

∴x= (5*50)/342

   =250/342

  = 1.368