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Given a rate of precipitation (in mm^3/hour), and average droplet diameter, is is it possible to calculate (or even ballpark) the density of the air+rain droplets at time of rainfall?

What I mean is, if I were to take a random metric cube of rain+air during rainfall, how much would it weigh?

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    $\begingroup$ The unit you suggest is not correct for rain rate, you would need to divide it by a surface area. That would give you for example mm/hour which equals L/m^2/hour. $\endgroup$ – Basileios May 10 at 20:22
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Air without rain has a mass of about $\ce{1200 g/m^3}$ at 20°C.

You can figure out the amount of water in falling rain if you know the rate of accumulation (how much rain falls through a given volume in a given time, landing on the ground) and the velocity of the falling rain.

Here's a link to a post elsewhere on this same question. In that discussion, they assume a rainfall rate of 1 inch (2.54cm) per hour, and estimate a raindrop's terminal velocity (from observation) at about 9 m/s. They divide that rate by that velocity (converting units), and come up with about $\ce{3/4 cm^3}$ of water per $\ce{m^3}$ of rain.

So, given those assumptions, the density of (rain + air) is about .06% higher than the density of air without rain. Interestingly, that's much less than the difference in density between dry and humidity-saturated air (about 1% at 20°C). In other words, air at 100% relative humidity with heavy rain falling through it is still less dense than dry air.

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You could calculate the mass of the water per cubic meter of air during rain by dividing the precipitation rate (in kg/m2/s) through the fall velocity (m/s) of the droplets. Or alternatively, from multiplying average droplet mass (volume x water density) with their number concentration.

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