Keosauqua, Iowa

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Keosauqua (pronounced /ˌkiː.əˈsɔːkwə/ KEE-ə-SAW-kwə) is a city in Van Buren County, Iowa, United States. The population was 1,066 at the 2000 census. It is the county seat of Van Buren County[1].

Contents

Geography

Keosauqua is located at 40°43′56″N 91°57′47″W / 40.73222°N 91.96306°W / 40.73222; -91.96306 (40.732089, -91.963027)[2].

According to the United States Census Bureau, the city has a total area of 1.6 square miles (4.1 km²), of which, 1.5 square miles (3.8 km²) of it is land and 0.1 square miles (0.3 km²) of it (7.59%) is water.

Demographics

As of the census[3] of 2000, there were 1,066 people, 467 households, and 270 families residing in the city. The population density was 728.7 people per square mile (281.9/km²). There were 505 housing units at an average density of 345.2/sq mi (133.5/km²). The racial makeup of the city was 98.59% White, 0.19% African American, 0.28% Asian, and 0.94% from two or more races. Hispanic or Latino of any race were 0.09% of the population.

There were 467 households out of which 24.6% had children under the age of 18 living with them, 44.5% were married couples living together, 10.5% had a female householder with no husband present, and 42.0% were non-families. 38.5% of all households were made up of individuals and 24.8% had someone living alone who was 65 years of age or older. The average household size was 2.10 and the average family size was 2.77.

In the city the population was spread out with 19.9% under the age of 18, 6.1% from 18 to 24, 19.5% from 25 to 44, 23.1% from 45 to 64, and 31.4% who were 65 years of age or older. The median age was 49 years. For every 100 females there were 77.4 males. For every 100 females age 18 and over, there were 73.9 males.

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