# In Henry law , Why mole fraction is equal mass of concentration (solution is given in pic)?

The partial pressure of ethane over a solution containing 6.56 x 10-3 g of ethane is 1 bar. If the solution contains 5.00 x 10-2 g of ethane, then what shall be the partial pressure of the gas?

### 1 Answer

- pisgahchemistLv 71 year ago
Partial pressure of ethane

It's a simple set of ratios:

Case 1 ... P = 1.00 bar, mass of solute = 0.00656g ethane

Case 2 ... P = ??? bar, mass of solute = 0.0500g ethane

P1 / mass 1 = P2 / mass 2

P2 = P1 x mass 2 / mass 1 = 1.00 bar x 0.0500g / 0.00656g = 7.62 bar

===========

Applying Henry's law:

c = kP ........ c = concentration, k = Henry's law constant, P = partial pressure

k = c / P

k = 0.00656g C2H8 x (1 mol C2H8 / 30.0g C2H8) / 1.00 bar

k = 0.0500g C2H8 x (1 mol C2H8 / 30.0g C2H8) / x bar

both are equal to k, so set them equal

0.00656g C2H8 x (1 mol C2H8 / 30.0g C2H8) / 1.00 bar = 0.0500g C2H8 x (1 mol C2H8 / 30.0g C2H8) / x bar

Cancel out the molar mass and solve for x

0.00656g C2H8 / 1.00 bar = 0.0500g C2H8 / x bar

x bar = 0.0500g / 0.00656g x 1.00 bar

x bar = 7.62 bar

And that's why we can use the ratio of the masses. The molar mass in the conversion to moles cancels out.

As for the answer you posted. The final answer is written to too many significant digits, and it probably contains some round-off error. It's a bit hard to read, so I'm not sure where the problem in it lies. But clearly, there is more than one way to solve the problem.

- Log in to reply to the answers