2 1/2 Cups of Raisins to Pounds Conversion

Question:
How many pounds of raisins in 2 1/2 US cups? How much are 2 1/2 cups of raisins in pounds?

The answer is:
2 1/2 US cups of raisins is equivalent to 0.876 ( ~ 1) pounds^(*)

Volume to 'Weight' Converter

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Results: 2 1/2 US cups of raisins equals 0.876 ( ~ 1) pounds. (*)
(*) To be more precise, 2 1/2 US cups of raisins is equal to 0.87627 pounds. All figures are approximate.

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US cups of raisins to pounds Chart

US cups of raisins to pounds
1.6 US cups of raisins	=	0.561 pounds
1.7 US cups of raisins	=	0.596 pounds
1.8 US cups of raisins	=	0.631 pounds
1.9 US cups of raisins	=	0.666 pounds
2 US cups of raisins	=	0.701 pounds
2.1 US cups of raisins	=	0.736 pounds
2 ¹/₅ US cups of raisins	=	0.771 pounds
2.3 US cups of raisins	=	0.806 pounds
2.4 US cups of raisins	=	0.841 pounds
2 ¹/₂ US cups of raisins	=	0.876 pounds

US cups of raisins to pounds
2 ¹/₂ US cups of raisins	=	0.876 pounds
2.6 US cups of raisins	=	0.911 pounds
2.7 US cups of raisins	=	0.946 pounds
2.8 US cups of raisins	=	0.981 pounds
2.9 US cups of raisins	=	1.02 pounds
3 US cups of raisins	=	1.05 pounds
3.1 US cups of raisins	=	1.09 pounds
3 ¹/₅ US cups of raisins	=	1.12 pounds
3.3 US cups of raisins	=	1.16 pounds
3.4 US cups of raisins	=	1.19 pounds

Note: some values may be rounded.

FAQs on raisins weight to volume conversion

2 1/2 US cups of raisins equals how many pounds?

2 1/2 US cups of raisins is equivalent 0.876 ( ~ 1) pounds.

How much is 0.876 pounds of raisins in US cups?

0.876 pounds of raisins equals 2 1/2 ( ~ 2 ¹/₂) US cups.

Weight to Volume Conversions - Cooking Ingredients

References:

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.