Diluting 100 ml. of rose water to 2.5% - Office Equipment & Supplies

not sure if this will help you out.
American Standard (Cups & Quarts ) American Standard (Ounces) Metric (Milliliters & Liters) 2 tbsp 1 fl. oz. 30 ml 1/4 cup 2 fl. oz. 60 ml 1/2 cup 4 fl. oz. 125 ml 1 cup 8 fl. oz. 250 ml 4 more rows

Equivalents and Measures ' Exploratorium

https://www.exploratorium.edu/cooking/convert/..
or try this link... How many cups of water is 250 ml Google Search

100 mils to 1 litre is 100 mils to 1000 mils or a ratio of 10:1 (1mil to every 10mils)

Concentration(start) x Volume(start) = Concentration(final) x Volume(final)

This equation is commonly abbreviated as: C1V1 = C2V2

Cleaning Instructions:
Water System
1) Dilute approximately 5 fl. oz. (148 ml) of recommended cleaner Hoshizaki " Scale Away" or "Lime-A-Way" ( Economics Laboratory, Inc) with 1 gal. (3.8 lit) of water.
2) Remove all ice from the Evaporator and the storage bin. Note: To remove cubes on the evaporator, remove the front panel ( Lower), move the control switch on the control box to the "OFF" position and move it back to the "ICE" position after 2 minutes. The harvest cycle starts and the cubes will be removed from the evaporator plate.
3) Move the control switch to the "OFF" position. Close the water supply line shut-off valve.
4) Remove the water tank to drain the water. Refit the water tank in its correct position.
5) Slowly pour the cleaning solution into the water tank.
6) Move the control switch to the "WASH" position.
7) After circulating the cleaner solution for about 30 minutes, move the control switch to the "OFF" position.
8) Remove the water tank to drain the cleaning solution. Refit the water tank in its correct position.
9) Open the shut-off-valve.
10) Move the control switch to the "WASH" position to circulate clean water for about 5 minutes.
11) Move the control switch to the "OFF" position and immediately back to the "WASH" position to rinse water.
12) Repeat the above rinse procedure three more times to rinse thoroughly.
13) Dilute approximately 0.5 fl. oz. (14.8 ml) of 5.25% sodium hypochlorite (Chlorine bleach) with 1 gal. (3.8 lit) of water.
14) Move the control switch to the "OFF" position. Close the water supply line Shut-off-valve.
15) Remove the water tank to drain the water. Refit the water tank in its correct position.
16) Slowly pour the sanitizing solution into the water tank.
17) Move the control switch to the "WASH" position.
18) After circulating the sanitizing solution for about 15 minutes, move the control switch to the "OFF" position.
19) Remove the water tank to drain the sanitizing solution. Refit the water tank in its correct position.
20) Open the shut-off valve.
21) Move the control switch to the "WASH" position to circulate clean water for about 5 minutes.
22) Move the control switch to the (OFF) position and immediately back to the "WASH" position to rinse water.
23) Repeat the above rinse procedure two more times to rinse thoroughly.
24) Move the control switch to the "ICE" position, and start the automatic ice making process. Refit the front panel (LOWER) in its correct position. abut do not forget to wash and sanitize the ice bin.

You cannot convert a unit of mass (mg) to a unit of capacity (volume). Mass and volume of a specific substance are related through the mass density (mass per unit volume).
What you have seems to be a dilution problem but it is not formulated completely. Please try again with necessary details.

Here is a quick conversion chart:

100 ml = 3.3 or 3.4 oz
50 ml = 1.7 oz
30 ml = 1 oz
15 ml = 1/2 oz
10 ml = 1/3 oz
7.5 ml = 1/4 oz
5 ml = 1/6 oz
3.7 ml = 1/8 oz = 1 dram

Note that all of these are approximate; strictly speaking, 1/2 oz is 14.787 ml.

A standard perfume sample is in a 1/32 oz (1 ml) vial.

This problem can be easily solved by using proportions:
To use the proportions you have to be aware of the meaning of percent (%).
Percent aways refers to how much of something compared to 100 of it.

For example, as in your problem, there are 95 grams of Na2CO3 per 100 mL of solution.

Percent, when applied to solutions, is more commonly a mass per mass ratio times 100; but for this problem, since you did not specify, I am going to assume you meant grams per volume, since the problem is asking for the "amount" of the Na2CO3 solution, which seems to be the solution volume.

That is, the unknown for this problem is the volume of the 95% Na2CO3.

You want to prepare 5 liters (i.e., 5 L) of a 0.5% solution of Na2CO3.
You don't say what the unit is for the 0.5, but I am taking the liberty of assuming you meant 0.5%.

Another way to state what you want to prepare is this:
You want to prepare 5 L of Na2CO3 solution in which there are 0.5 grams of Na2CO3 per 100 mL of solution.

The percent ratio can be used as a conversion factor, which can be written as a ratio:
0.5 g Na2CO3 / 100 mL solution.
It is important to use the units (in boldface) for each numerical value, because they will guide you error-free in this type of problem! Remember, you can write it this way, because % in this problem means how much (in grams) per 100 (in mL).

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OK, here is how you use this conversion factor: You multiply it by the given quantity, which is 5 L or 5,000 mL. (I converted the L to mL by multiplying it by 1000, since there are 1000 mL in 1 L - it's a good idea to memorize that.)
5000 mL x (0.5 g / 100 mL).
Remember, always multiply the given quantity (not a ratio) by the conversion factor (which is a ratio). You can assure accuracy by being sure that the like-units (mL) cancel out.
The calculated result of this math operation is 25 g (of the solute, Na2CO3).
Note that the unit for this result is g (or grams) of Na2CO3; you should always be sure to attach the complete unit to the numerical value 25. Also note that the calculated result is actually only a one-sigfig quantity; I indicated that by putting the sig fig in boldface and also underlining it.

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Before you can complete this problem, you need to use one more conversion factor, already mentioned in the one-sentence 2nd paragraph above:
95 g Na2CO3 / 100 mL solution
However, you will have to invert or flip it as follows:
100 mL solution / 95 g Na2CO3
You will use it as above for the first conversion factor; you will multiply it by a different quantity (not a ratio), as follows:
25 g Na2CO3 x (100 mL solution / 95 g Na2CO3)
Do you see why I flipped the ratio now? To allow cancelling out of the like units, "g Na2CO3."

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So, the answer to this problem is 26.32 mL of the 95% Na2CO3 solution (before proper rounding off).

After rounding off using the rules for handling sig figs,
the correct answer is 30 mL of the 95% solution of Na2CO3.

Additional Notes
Some information about rounding off
Notice that in the above paragraph, and in the two paragraphs before that, I have underlined the only digit that is considered to be a significant digit (sig fig). An important rule to remember regarding sig figs and rounding off in math operations involving muliplication or division is this:
The calculated result may not have more sig figs in it than was in the quantity with the smallest number of sig figs. So there can be only one sig fig in 26.32, because it was calculated using another one-sigfig number, 25. (The 25 was obtained from calculation using another one-sigfig number, 0.5). To learn more about sig figs and how to consider them properly when rounding off, you may consult almost any chemistry textbook, or Google up on the world wide web.

Regarding the question implied by the problem, it is not based on a realistic situation. The reason I say this is because there is no such thing as a 95% aqueous solution of Na2CO3. Only about 25 grams will dissolve in 100 mL of water. (This fact may be confirmed by looking up the solubility of Na2CO3 in water.)

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100 ml of 2cycle oil to 4000 ml of fuel..Which is 40/1 ratio
100 ml of 2cycle oil to 5000 ml of fuel.. which is 50/1 ratio

4 o/z is 118.29411875006815 ml to 1 gallon of fuel (3.79 litres or 3790 ml) Which means 32/1 ratio...I find that to confusing,so it is easier to say 100 ml of oil to 3200 ml of fuel ===32/1.

Hello, I can help you if you include more data in your question.

As an organic compound, lidocaine is insoluble in water, so I am assuming that either some combination of water and alcohol (ethanol), or a pure organic solvent like ethanol or diethyl ether is being used.

If lidocaine were soluble in water (which it's not), you would have been able to solve your problem as indicated below. Maybe you would be able to modify the indicated solution by taking into consideration the density of the organic solvent.

Percentage (%) by mass is one of the common ways to express the concentrations of solutions.
Since % means the number of grams of solute in 100 grams of solution (solute + solvent), a 1.5% solution of any solute refers to 1.5 grams of it in 100 grams of solution.

Assuming we are referring to an aqueous solution (that is, where water is the solvent), this reasonably approximates to 1.5 grams of lidocaine per 100 mL solution. We can assume this result, because we have a dilute watery solution, which probably has the density close to the density of pure water, which is 1.0 gram per 1.0 mL at commonly used temperatures, such as average room temperature.

So, 1.5 g lidocaine per 100 mL solution can easily be scaled down by one-tenth to give:
0.15 g lidocaine per 10 mL solution.

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In a properly working machine you should get about 50 ml (1.5 ounces) for the single and about 100 ml for the double. The amount is also programmable and just about the only way to get the same amount is to program both to deliver the same amount. The unit has a flow meter to accurately measure the amount of water going to the espresso. Try the program function to reset the amount for single and double.