Question

ion
7g of a solute are dissolved in 2009 of solvent (mol.wt. =78) to prepare a solution AL ACE
temperature dry air is sent into a solution and solvent continuously. The loss in weights of
containers containing solution and solvent are 0.975 g and 0.025g. Calculate the molama
of solute.
Lowering of vapour pressure (Po-P) is proportional to y; y = 0.025g
Vapour pressure (P) is proportional to x:x= 0.975 g
Vapour pressure of pure solvent = P = x + 4 = 0.975 +0.025 = 1g
y po-p_WM 0.025
x+y po Wm - 1.0
7478
Molar mass of solute, m=-
= 100g mol-1
0.025 200​

Answer 1

Answer:

Explanation:

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