The boiling point of isopropyl alcohol is 82.5°C. a student conducts an experiment and finds a boiling point of 83.1°C. what is the students percent error in this experiment? A.0.6%
B. 0.7%
C. 1.0%
D. 99.3%

B. 0.7%

Explanation:

Given data:

Actual boiling point = 82.5°C

Experimental boiling point = 83.1°C

Percent error = ?

Solution:

Formula:

Percent error = ( actual value - experimental value / actual value )×100

by putting values,

Percent error = (82.5°C - 83.1 °C /82.5°C) × 100

Percent error = 0.007 × 100

Percent error = 0.7 %

Negative sign shows that experimental value is greater than accepted value. It can not written in result.

the answer is b. dilution