# Question #d1256

Jul 7, 2017

Density of the final mixture is 1133 $\frac{k g}{m} ^ 3$

#### Explanation:

You need to compute final density. When you mix two different compounds:

$\left({V}_{1} \times {d}_{1}\right) + \left({V}_{2} \times {d}_{2}\right) = \left({V}_{1} + {V}_{2}\right) \times {d}_{m i x t u r e}$

$\left(1 \times 1\right) + \left(2 \times 1.2\right) = 3 \times {d}_{m i x t u r e}$

Note that 1000 $\frac{k g}{m} ^ 3$ is the same as 1 kg per liter. The density of brine is therefore 1.2 $\frac{k g}{L}$.

Final volume is 3 liters because you indicated that there is no volume change when these two are mixed well.

You can solve the above equation:

$1 + 2.4 = 3 \times {d}_{m i x t u r e}$

$\frac{3.4}{3} = {d}_{m i x t u r e}$

${d}_{m i x t u r e} = 1.133 \frac{k g}{L}$

or density of the final mix is 1133 kg per cubic meter.