What weight of non-volatile solute urea ( $\mathop {NH}\nolimits_2 - CO - \mathop {NH}\nolimits_2 $) needs to be dissolve in 100gram of water by 25% ( in gram)

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Hint: When a non-volatile solute is added to a solvent , the vapour pressure of the solution becomes less than that of the solvent it is roult’s law . This occurs in case of relative lowering of vapour pressure. According to roult’s law $p^\circ A - \dfrac{{\mathop p\nolimits_S }}{{p^\circ A}}$ = relative lowering in vapour pressure = $\mathop X\nolimits_B $
Where $p^oA$ is vapour pressure of pure solvent, Ps is vapour pressure of solution and $X_B$ is mole fraction of solute .

Complete step by step answer:
Consider vapour pressure of water = 100
So vapour pressure of urea solution = 100-25 = 75
$p^\circ A - \dfrac{{\mathop p\nolimits_S }}{{p^\circ A}}$= \[\dfrac{{\mathop N\nolimits_2 }}{{\mathop N\nolimits_1 + \mathop N\nolimits_1 }}\]($\mathop N\nolimits_2 = \dfrac{{100}}{{18}}$)------eqn1
In equation1
$\mathop W\nolimits_1 $= weight of water = 100g
$\mathop W\nolimits_2 $ = weight of urea that we have to find...
$\mathop M\nolimits_1 $= molecular weight of urea = 60g
$\mathop M\nolimits_2 $ = molecular weight of water = 18g
Now putting these values in equation 1
$100 - \dfrac{{75}}{{100}} = \dfrac{{\mathop N\nolimits_2 }}{{\dfrac{{100}}{{18}} + \mathop N\nolimits_2 }}$ ($\mathop N\nolimits_2 = \dfrac{{\mathop W\nolimits_2 }}{{60}}$)
$\mathop W\nolimits_2 $= 111.1g

Measurement of relative lowering of vapour pressure can be found if pure dry air is passed through bulbs containing solution and then through bulbs containing pure solvent and finally through tubes containing anhydrous $CaC\mathop l\nolimits_2 $.