Salt water seed selection is a kind of seed selection method invented by the working people in China. At present, salt water with density of 1.1 × 10 ~ 3 kg / M ~ 3 is needed. If the quality of 0.6 DM3 salt water is 540 G (1) Please calculate whether the brine meets the requirements? (2) If not, should salt or water be added? How many kg?

Salt water seed selection is a kind of seed selection method invented by the working people in China. At present, salt water with density of 1.1 × 10 ~ 3 kg / M ~ 3 is needed. If the quality of 0.6 DM3 salt water is 540 G (1) Please calculate whether the brine meets the requirements? (2) If not, should salt or water be added? How many kg?

At present, the prepared brine density is 540g / 0.6dm = 900kg / M ~ 3, which is less than the required density. Therefore, salt should be added: 0.6dm ~ 3 * 1.1 * 10 ~ 3kg / M ~ 3-540g = 0.12kg

When selecting seeds with salt water, the density of salt water is 1.1 * 10 ^ 3kg / m ^ 3. Now 350ml salt water is prepared, and the weight is 0.6kg. How much water should be added to meet the standard?
I have calculated that the density of 350 ml brine with a mass of 0.6 kg is 1.7 g / cm ^ 3

I've thought about it, but I don't know. I don't know the concentration of salt water in this question. I can't work it out

The density of salt water is 1.1 × 103 kg / m3. The salt water of 0.5 decimeter 3 is prepared, and its weight is 0.6 kg,
Select seeds with salt water, the density of salt water is 1.5 × 103 kg / m3, make 0.5 decimeter 3 salt water, and weigh it to be 0.6 kg. Ask whether the salt water meets the requirements? If not, add water or salt? If not, how much?
Salt water density is required to be 1.1 × 103 (wrong number)

Density = m / v = 0.6 / 0.5 * 10 ^ - 3 = 1.2 * 103
So the density is too high, we need to add water
If M kg of water is added, there will be m / 103 cubic meters of water added, so it can be substituted
1.1*103＝（m+M）/（M/10^3+0.5*10^-3）
The solution is m = 0.1kg
Feel right, please give points, thank you