well its back-of-the-envelope time:
(a) assuming acetone was at its odour threshold (~150 ppm)
(b) assuming the pond was 200 m away
(c) assuming no acetone remains unvolatilised
(c) in a complete bastardisation of everything we know about air density, wind and diffusion, assuming the concentration of acetone is even over a hemisphere centred over the pond, with its outer edge at the observer
So volume of a sphere is 4/3.pi.r^3, which gives our volume of the hemisphere as 16.74 x10^6 cubic metres; which means there is 16.74 x10^3 L of space inside the hemisphere. So if all of the air inside that space is at 150 ppm of acetone, then there is 2511 litres of gaseous acetone in that space. From the ideal gas law we know the volume of a gas is 22.4 L/mol, so there is 112 moles of acetone in the hemisphere, or about 6.5 kg (or about 8.2 L).
Note the reliance on a series of estimates and approximations - some of them quite major. This number is not to be taken as gospel, but its an estimate. And it shows that comparatively little acetone is required to produce a smell over a wide area.