Question

9) After lab, all of Darrel’s friends looked at his data and laughed and laughed. They told him that he was 30.8% too low in the boiling point he had just recorded. He had recorded a boiling point of 50o C on his data sheet. What is the correct boiling point of the liquid he was working with in lab?

Answer 1

Answer:

$\boxed{72 \, ^{\circ}\text{C}}$

Explanation:

$\text{Percent error} = \dfrac{\lvert \text{Measured - Actual}\lvert}{ \text{Actual}} \times100 \,\%$

Data:

Percent error = 30.8 %

    Measured = 50 °C

Calculation:

Let x = the actual value

$\begin{array}{rcl}-30.8 \,\% & = &\dfrac{50 - x}{x} \times 100 \, \%\\\\-30.8x & = & 100(50 - x)\\-30.8x & = & 5000 - 100x\\69.2x & = & 5000\\x & = & \dfrac{5000}{69.2}\\ & = & \mathbf{72}\end{array}$

$\text{The correct boiling point is }\boxed{\mathbf{72 \, ^{\circ}\text{C}}}\\\text{Note: The answer can have only two significant figures, because that is all} \\\text{you gave for the measured value.}$

Answer 2

If he was 30.8% too low, it means that he was at 69.2% of the boiling point needed. So 50o C is 69.2% of total.

In order to know what 100% is, you can divide the number by it's percentage and then multiply it by a hundred.

So: 50/30.8=1.623
1.623*100=162.3
So the correct boiling point of the liquid he was working with in the lab is 162.3 oC