Question

a student is testing the boiling point of an unknown liquid. he calculated a 8.0% error. the theoretical boiling point of the unknown liquid should be 75°c. what is the experimental boiling point of his unknown liquid?

Explanation:

Step1: Recall the formula for percent error

The formula for percent error is $\text{Percent Error} = \frac{|\text{Experimental Value} - \text{Theoretical Value}|}{\text{Theoretical Value}} \times 100\%$. We know the percent error is $8.0\%$ and the theoretical value ($T$) is $75^\circ\text{C}$. Let the experimental value be $E$. So, $8.0\%=\frac{|E - 75|}{75}\times 100\%$.

Step2: Solve for the absolute difference

First, divide both sides by $100\%$: $0.08=\frac{|E - 75|}{75}$. Then multiply both sides by $75$: $|E - 75|=0.08\times75 = 6$.

Step3: Find the experimental value

This gives two cases: $E - 75 = 6$ or $E - 75=-6$.

• For $E - 75 = 6$, $E=75 + 6=81^\circ\text{C}$.
• For $E - 75=-6$, $E=75-6 = 69^\circ\text{C}$. But since percent error is positive and we are looking for the experimental value, we consider the magnitude. However, usually, if the error is positive (overestimation) or negative (underestimation), but since the problem says "experimental boiling point", we can calculate both, but likely the student's experimental value is either $81^\circ\text{C}$ (if the error is due to overestimation) or $69^\circ\text{C}$ (if due to underestimation). But let's check the formula again. Wait, the percent error formula is $\text{Percent Error}=\frac{|\text{Exp}-\text{Theo}|}{\text{Theo}}\times100\%$. So we can also re - arrange for $\text{Exp}=\text{Theo}\times(1\pm\frac{\text{Percent Error}}{100})$.

Using the positive sign (overestimation): $\text{Exp}=75\times(1 + 0.08)=75\times1.08 = 81^\circ\text{C}$.
Using the negative sign (underestimation): $\text{Exp}=75\times(1 - 0.08)=75\times0.92 = 69^\circ\text{C}$. But the problem is asking for the experimental boiling point. Since the student calculated an $8.0\%$ error, we need to see if it's a positive or negative error. But since the problem doesn't specify over or under, but in most cases, if we assume the error is calculated as $\frac{\text{Exp}-\text{Theo}}{\text{Theo}}\times100\%$ (without absolute value, but usually percent error uses absolute value). Wait, the problem says "8.0% error", so using the absolute value formula, we have two possibilities. But maybe the question is assuming that the experimental value is higher or lower. But let's go with the formula. Let's assume that the error is $\frac{\text{Exp}-\text{Theo}}{\text{Theo}}\times100\%=8.0\%$ (so overestimation). Then $\text{Exp}=\text{Theo}\times(1+\frac{8}{100})=75\times1.08 = 81^\circ\text{C}$. Or if it's underestimation, $\text{Exp}=75\times(1 - 0.08)=69^\circ\text{C}$. But since the problem is about a student's experimental value, maybe we can calculate both, but likely the intended answer is using the formula $\text{Exp}=\text{Theo}+\text{Theo}\times\frac{\text{Percent Error}}{100}$ (if error is positive) or $\text{Exp}=\text{Theo}-\text{Theo}\times\frac{\text{Percent Error}}{100}$ (if negative). But let's check the calculation again.

Answer:

If we assume the error is a positive error (experimental value is higher than theoretical), the experimental boiling point is $\boldsymbol{81^\circ\text{C}}$. If we assume negative error, it is $\boldsymbol{69^\circ\text{C}}$. (Most likely, if we consider the percent error formula with the absolute value and solving for experimental value, and if we take the case where the experimental value is higher, the answer is $81^\circ\text{C}$)