What will be the mass of a solution that has a density of "1/15 g/mL" and a volume of "2.00 L" ?

Sep 20, 2015

$2.30 \cdot {10}^{3} \text{g}$

Explanation:

You know that density is defined as mass per unit of volume.

In your case, you know that your solution has avolume of $\text{2.00 L}$ and a density of $\text{1.15 g/mL}$.

The first thing to notice here is that the volume of the solution is given to you in liters, but that the density is expressed in grams per mililiter.

At this point, you have a choice. You can either convert the solution's density from grams per liter to grams per liter, or even kilograms per liter, or convert the volume of the solution to mililiters.

I'll show you one approach, and you can use the other one as practice.

So, let's say that you want to use the volume in liters. Convert the density from grams per mililiter to grams per liter

$1.15 \text{g"/color(red)(cancel(color(black)("mL"))) * (1000color(red)(cancel(color(black)("mL"))))/"1 L" = "1150 g/L}$

The mass of the solution will thus be

2.00color(red)(cancel(color(black)("L"))) * "1150 g"/(1color(red)(cancel(color(black)("L")))) = color(green)(2.30 * 10""^3"g")

If you want, you can express this in kilograms

2.30 * color(red)(cancel(color(black)(10^3)))color(red)(cancel(color(black)("g"))) * "1 kg"/(color(red)(cancel(color(black)(10^3)))color(red)(cancel(color(black)("g")))) = "2.30 kg"

The answer must be rounded to three sig figs, the number of sig figs you gave for the volume and density of the solution.

Sep 20, 2015

$2.3 \text{kg}$

Explanation:

$\rho = \frac{m}{v}$

$m = \rho \times v$

$1.15 \text{g/ml"=1.15xx10^(3)"g/l}$

So:

$m = 1.15 \times {10}^{3} \times 2 = 2.3 \times {10}^{3} \text{g}$

$m = 2.3 \text{kg}$