# 2.4   Calculate the mass of urea  $\inline (NH_{2}CONH_{2})$  required in making  $\inline 2.5 \; kg$  of  $\inline 0.25\; molal$  aqueous solution.

Let us assume that the mass of urea required be x g.

So moles of urea will be :

$Moles = \frac{Given\ mass}{Molar\ mass} = \frac{x}{60}\ moles$

$Molality = \frac{Moles}{Mass\ of\ solvent\ in\ Kg} = \frac{\frac{x}{60}}{2.5-0.001x} = 0.25$

we get      x =  37

Thus mass of urea required = 37 g.

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