Orderville is a town in Kane County, Utah, United States. The population was 596 at the 2000 census.
Orderville is located at 37°16′33.39″N 112°37′56.89″W / 37.2759417°N 112.6324694°W / 37.2759417; -112.6324694 (37.262348, -112.653266).
According to the United States Census Bureau, the town has a total area of 9.2 square miles (23.8 km²), all of it land.
As of the census of 2000, there were 596 people, 194 households, and 150 families residing in the town. The population density was 64.9 people per square mile (25.1/km²). There were 240 housing units at an average density of 26.1 per square mile (10.1/km²). The racial makeup of the town was 97.65% White, 0.50% Native American, 0.17% from other races, and 1.68% from two or more races. Hispanic or Latino of any race were 1.51% of the population.
There were 194 households out of which 39.2% had children under the age of 18 living with them, 67.0% were married couples living together, 7.7% had a female householder with no husband…
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