Solving Density Problems

Solving Density Problems-18
After some additional time has passed, 15 liters of gas is used up and a fresh 15 liters of gas is pumped into the tank to replace it.

After some additional time has passed, 15 liters of gas is used up and a fresh 15 liters of gas is pumped into the tank to replace it.

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This corresponds to 1/2 of the stabilizer remaining, which is 66.67×1/2 = 33.33 m L.

We now add 15 liters of gas to the tank, bringing it back up to 30 liters, but with only 33.33 m L of stabilizer left.

Density is the measurement of the amount of mass per unit volume.

Density calculations are done using the formula: where ρ = density m = mass V = volume Example Problems: 1.

Next, 10 liters of fresh gas is added, but no additional stabilizer is added, hence the amount of stabilizer remains at 66.67 m L (with 30 liters of gas now inside the tank).

Next, 15 liters of gas is used up, which translates into 15 liters left in the tank, or 1/2 of the fuel remaining.Equate the mass of the cylinder to the mass of the water displaced by the cylinder. For the cylinder, volume = (cross-sectional area) × length. The cross-sectional area cancels out and we can easily calculate the density of the cylinder.The density is 0.7 g/cm This is a nice variation on the usual density problems.Assume the volume of the two liquids is additive when mixed.Hint and answer Problem # 7 It is given that 10 m L of fuel stabilizer treats 3 L of gasoline.It involves looking at the problem from the point of view of concentration.The initial amount of stabilizer in the tank is 100 m L.We know that for 30 liters of fuel we must add 100 m L of stabilizer.Hence, the amount of stabilizer to add is 100−33.33 = 66.67 m L.After 10 liters of gas is used up the first time, we have 20 liters remaining inside the tank.This corresponds to 2/3 of the fuel remaining, which also corresponds to 2/3 of the stabilizer remaining, which is 100×2/3 = 66.67 m L.


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