2 2/3 Cups of Polenta to Pounds Conversion

Question:
How many pounds of polenta in 2 2/3 US cups? How much are 2 2/3 cups of polenta in pounds?

The answer is:
2 2/3 US cups of polenta is equivalent to 0.94 ( ~ 1) pound(*)

Volume to 'Weight' Converter

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Results:

2 2/3 US cups of polenta equals 0.94 ( ~ 1) pound. (*)
(*) To be more precise, 2 2/3 US cups of polenta is equal to 0.94037 pound. All figures are approximate.

US cups of polenta to pounds Chart

US cups of polenta to pounds
1.767 US cup of polenta = 0.623 pound
1.867 US cup of polenta = 0.658 pound
1.967 US cup of polenta = 0.694 pound
2.067 US cups of polenta = 0.729 pound
2.167 US cups of polenta = 0.764 pound
2.267 US cups of polenta = 0.799 pound
2.367 US cups of polenta = 0.835 pound
2.467 US cups of polenta = 0.87 pound
2.567 US cups of polenta = 0.905 pound
2.67 US cups of polenta = 0.94 pound
US cups of polenta to pounds
2.67 US cups of polenta = 0.94 pound
2.767 US cups of polenta = 0.976 pound
2.867 US cups of polenta = 1.01 pound
2.967 US cups of polenta = 1.05 pound
3.067 US cups of polenta = 1.08 pound
3.167 US cups of polenta = 1.12 pound
3.267 US cups of polenta = 1.15 pound
3.367 US cups of polenta = 1.19 pound
3.467 US cups of polenta = 1.22 pound
3.567 US cups of polenta = 1.26 pound

Note: some values may be rounded.

FAQs on polenta weight to volume conversion

2 2/3 US cups of polenta equals how many pounds?

2 2/3 US cups of polenta is equivalent 0.94 ( ~ 1) pound.

How much is 0.94 pound of polenta in US cups?

0.94 pound of polenta equals 2 2/3 ( ~ 2 3/4) US cups.

Weight to Volume Conversions - Cooking Ingredients

References:

  1. Density Database Version 2.0 - FAO

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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