At 25 °C, the vapour pressure of pure water is 23.76 mm Hg and that of an aqueous dilute solution of urea is 22.98mm Hg. Calculate the molality of this solution.
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Answered by
16
At 25°C, urea is in the form of a non volatile solute.
∴ According to Raoult's law for non volatile solute;
⇒
Now, we know that,
The mole fraction of water = 0.9875
∴ the mole fraction of urea = (1 - 0.9875) = 0.0324
Thus, the Molality of this solution can be calculated as,
= m
= 1.82 m
Ans) The molality of the given solution under the specified temperature is 1.82 Mol
Answered by
10
molality is 0.968m under following conditions.
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