1 Ml of Vinegar to Pounds Conversion
Question:
How many pounds of vinegar in 1 milliliter? How much is 1 ml of vinegar in pounds?
The answer is:
1 milliliter of vinegar is equivalent to 0.00214 pound(*)
Volume to 'Weight' Converter
Milliliters of vinegar to pounds Chart
Milliliters of vinegar to pounds | ||
---|---|---|
0.1 milliliter of vinegar | = | 0.000214 pound |
1/5 milliliter of vinegar | = | 0.000429 pound |
0.3 milliliter of vinegar | = | 0.000643 pound |
0.4 milliliter of vinegar | = | 0.000857 pound |
1/2 milliliter of vinegar | = | 0.00107 pound |
0.6 milliliter of vinegar | = | 0.00129 pound |
0.7 milliliter of vinegar | = | 0.0015 pound |
0.8 milliliter of vinegar | = | 0.00171 pound |
0.9 milliliter of vinegar | = | 0.00193 pound |
1 milliliter of vinegar | = | 0.00214 pound |
Milliliters of vinegar to pounds | ||
---|---|---|
1 milliliter of vinegar | = | 0.00214 pound |
1.1 milliliter of vinegar | = | 0.00236 pound |
1 1/5 milliliter of vinegar | = | 0.00257 pound |
1.3 milliliter of vinegar | = | 0.00279 pound |
1.4 milliliter of vinegar | = | 0.003 pound |
1 1/2 milliliter of vinegar | = | 0.00321 pound |
1.6 milliliter of vinegar | = | 0.00343 pound |
1.7 milliliter of vinegar | = | 0.00364 pound |
1.8 milliliter of vinegar | = | 0.00386 pound |
1.9 milliliter of vinegar | = | 0.00407 pound |
Note: some values may be rounded.
FAQs on vinegar weight to volume conversion
1 milliliter of vinegar equals how many pounds?
1 milliliter of vinegar is equivalent 0.00214 pound.
How much is 0.00214 pound of vinegar in milliliters?
0.00214 pound of vinegar equals 1 milliliter.
Weight to Volume Conversions - Cooking Ingredients
References:
Notes on ingredient measurements
It is a bit tricky to get an accurate food conversion since its characteristics change according to humidity, temperature, or how well packed the ingredient is. Ingredients that contain the terms sliced, minced, diced, crushed, chopped add uncertainties to the measurements. A good practice is to measure ingredients by weight, not by volume so that the error is decreased.
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