How to Calculate the Mass of Benzene at 15 Degrees Celsius?

[MacLaddy]

Homework Statement

The density of Benzene at 15 degrees Celsius is $0.8787 \frac{g}{mL}$ Calculate the mass of 0.1500 L of Benzene at this temperature.

Homework Equations

$Density=\frac{mass}{volume}$

The Attempt at a Solution

The part of this problem that is throwing me off is the g/mL, and L measurements. What is have done is I took the density and multiplied it by 1000 to get g/L instead of g/mL. The equation then looked like this.

$878.7\frac{g}{L} = \frac{mass}{0.1500L}\rightarrow 878.7(0.1500)=mass\rightarrow 131.805g = mass$

I'm not sure if I am correct with that result, but even if I am I believe I am making a mistake with how I distribute L and g/mL across my multiplication. Any advice on this problem would be appreciated.

Mac

 

[Borek]

You did OK. You could as well convert volume to mL (0.1500L = 150.0mL) - final result would be identical.

 

[MacLaddy]

Thanks for the double-check, Borek.

I realized what was confusing me so badly. I knew that if I multiplied L by 1000 I could get mL, yet I was seeing that I also needed to multiply my g/mL by 1000 to get g/L. I was thinking, "how can I multiply both by 1000, shouldn't I be dividing one of the figures?" Anyhow, it finally dawned on me, it's g/mL, not just mL.

Thanks again.
Mac

 

[Borek]

I knew that if I multiplied mL by 1000 I could get L

Quite the opposite - mL has to be divided by 1000 to be converted to L. Using dimensional analysis:

$$10\ \text{mL} = 10\ \text{mL}\frac{1\ \text{L}}{1000\ \text{mL}} = 0.01\ \text{L}$$

 

[MacLaddy]

You're correct, Borek, and I misspoke. I was actually referring to multiplying L by 1000, as in your post here,

You did OK. You could as well convert volume to mL (0.1500L = 150.0mL) - final result would be identical.

I've corrected the error in that post.