6.970 g of a non-volatile solute is dissolved in 230.0 g of water.
The solute does not react with water nor dissociate in solution.
Assume that the resulting solution displays ideal Raoult's law behaviour.
At 60°C the vapour pressure of the solution is 147.48 torr.
The vapour pressure of pure water at 60°C is 149.40 torr.
Calculate the molar mass of the solute (g/mol
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Given:
Solution follow ideal Raoult's law behaviour
Solute = 6.970 g
Solvent (water) = 230 g
Solution pressure at 60°C = 147.8 torr
Pure water pressure at 60°C = 149.40 torr
To Find:
The molar mass of solute =?
Solution:
We know that by Raoult's law we get the relation,
---------(1)
here, vapour pressure of the solution
= mole fraction of solvent
= vapour pressure of the pure solution
We know the values,
147.48
= 149.40
=
Now, on substituting values in the equation-(1)
On solving for y
y = 30.30 g/mol
Thus, the molar mass of the solute is 30.30 g/mol
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