Question

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​

Answer 1

Answer:

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Explanation:

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Answer 2

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,

$P_{solution} = X_{solvent} P_{solvent}$        ---------(1)

here, $P_{solution} =$ vapour pressure of the solution

         $X_{solvent}$ = mole fraction of solvent

         $P_{solvent}$ = vapour pressure of the pure solution

We know the values,

$P_{solution} =$  147.48

$P_{solvent}$ = 149.40

$X_{solvent} = \dfrac{n_{water} }{n_{water} + n_{solute}}$

             = $\dfrac{\dfrac{230}{18} }{\dfrac{230}{18} + \dfrac{6.970}{y}}$

Now, on substituting values in the equation-(1)

$147.48 = \dfrac{\dfrac{230}{18} }{\dfrac{230}{18} + \dfrac{6.970}{y}} \times149.40$

On solving for y

$\dfrac{147.48}{149.40} = \dfrac{\dfrac{230}{18} }{\dfrac{230}{18} + \dfrac{6.970}{y}}$

$0.98 = \dfrac{12.77 }{12.77 + \dfrac{6.970}{y}}$

$13 = 12.77 + \dfrac{6.970}{y}$

y = 30.30 g/mol

Thus, the molar mass of the solute is 30.30 g/mol