Calculation of acraldehyde liquid phase concentration

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Acraldehyde is a liquid at room temperature and has a foul odor. Its molecular formula is C3H4O, and its CAS number is 107-02-8. Its boiling point is 52.7 degrees, its specific gravity is 0.84 at 20 degrees, and its vapor pressure is 29.3 KPa at 20 degrees. Trace amounts of acraldehyde are produced during the production process (as a result of the decomposition of glycerol), and it has an extremely unpleasant odor. Using a gas detection tube at the gas-liquid separation tank of the water ring vacuum pump, the acraldehyde content in the exhaust gas was measured to be 20 ppm. I would like to estimate how much acraldehyde is present in the water inside the water ring pump (with the water temperature at 20 degrees). Assuming an ideal solution, I calculated it using Raoult’s law as follows: The vapor pressure of acraldehyde = 101 KPa * 20 ppm/10^6 = 0.00202 KPa. The vapor pressure of acraldehyde at 20 degrees is 29.3 KPa. The mole fraction of acraldehyde in the liquid phase = 0.00202/29.3 = 6.9E(-5). Is this calculation correct? Please give me your advice. Thank you