Question

An air-dry soil with a moisture content of 35% by weight is to be used in a pot

experiment. If each pot is to have 20 kg oven-dried soil, how much of the air-dry soil should

be weighed into the pot and how much is the moisture of that soil in kg?

Answer 1

$m_w$ =water mass

$m_t$ = total mass

$(20+m_w) = m_t \\ m_w = 0.35m_t \\ 20 + 0.35m_t = m_t \\ 20 = 0.65m_t \\ m_t = 30 .77 \;kg$

30.77 kg of the air-dry soil should be weighed into the pot.

$m_w = 0.35 \times 30.77= 10. 77 \;kg$

10.77 kg of that is moist