Question

An aqueous solution containing 5.00% by mass of a non-volatile, non-dissociating organic compound had exactly the same vapor pressure at 298K as an aqueous solution containing 1.00 % by mass of NaCl. What was the molar mass of the organic compound.

Consider Henry's Law. Remember that NaCl dissociates in water

Answer 1

1% NaCl solution = 1gm NaCl is present in 100gm of water = 1gm NaCl in 100mL water

Moles of NaCl = 1gm/58.5gm/mol = 0.017 moles in 100mL water

Molarity = 0.017moles*1000mL/100mL = 0.17 M

osmotic pressure = iM RT

2 * 0.17 Moles/L * 0.082 Latm/k/mol *298 K = 8.31 atm

i=2 for naCl as NaCl dissociates in solution to give two ions.

Unknown compound has same osmotic pressure.

8.31 atm = i*M*R*T

or, 8.31 atm = 1*M*0.082 Latm/k/mol *298 K

M = 0.34

0.34 moles of unknown is present in 100mL water.

Now, according to the given problem, 5 gm compound is present in 100gm water.

So, 1000gm (=1000mL) water contains 50 gm compound

Moles of the unknown = 50gm/MW

50gm/MW = 0.34 moles

MW = 147.06 gm/mol

Molecular weight of the unknown is 147.06 gm/mol