Question

10% aqueous solution of sugar cane is isotonic with 1.745% aqueous solution of urea. find the molecular mass of urea.​

Answer 1

The molecular mass of Urea = $59.679 gm\ mol^{-1}$

Explanation:

Non-ideal solutions are those that do not follow Raoult's law. Colligative qualities are characteristics of a non-ideal solution that are dependent on the number of solute particles dispersed in a solvent.

The following formula is used to compute urea's molecular weight:

$\pi = iCRT\\\pi\ is\ the\ osmotic\ pressure\\i = vant\ Hoff\ factor\\C = molar\ concentration\ of\ the\ solute\ in\ the\ solution\\R = universal\ gas\ constant\\T = temperature$

Putting the osmotic pressure of the two solutions equal

$\pi _{1} = \pi _{2}\\i_{1} C_{1}RT = i_{2} C_{2}RT$

R and T are constant for both solutions.

$C_{1}i_{1} = C_{2}i_{2}$

Since both cane sugar and urea do not dissociate; $i_{1} =i_{2}$

$C = \frac{given weight}{Molecular\ weight}$

$\frac{Given\ weight\ of\ urea}{Molecular\ weight\ of\ urea} = \frac{Given\ weight\ of\ cane\ sugar}{Molecular\ weight\ of\ cane\ sugar}$

$Given\ weight\ of\ urea =$ 1.745% = $\frac{1.745}{100} gms$

$Given\ weight\ of\ cane\ sugar=$ 10% = $\frac{10}{100} gms$

$\\Molecular\ weight\ of\ cane\ sugar =$ $342 gm/mol^{-1}$

$\\Molecular\ weight\ of\ urea =$ X

$\frac{10}{100*342} = \frac{1.745}{100* X}$

$X= \frac{1.745*100*342}{100*10}$

$X = 59.679 gm\ mol^{-1}$