At $ 20^oC $ we mix $ 40.00 \; mL $ of ethanol at $ 60.00; mL $ of water. Calculate using → tables the masses of ethanol $(m_1)$ and water $(m_2)$ !
For answers, use (possibly multiple times) the arrows ↑ and ↓ at the top! Complete please this question before moving on to the next!
$m_1$ $=$ $\rho \cdot V$ $=$ $0,78945\cdot 40,00$ $ =$ $31,58\; g$ $m_2$ $=$ $\rho \cdot V$ $=$ $0,998203\cdot 60,00$ $ =$ $59,89\; g$
Then calculate the mass of the mixture !
$m_{12}$ $=$ $m_1+m_2$ $=$ $91,47\; g$
Then calculate the mass percentage of ethanol !
$\%$ $=$ $\frac{m_1\cdot 100}{m_{12}}$ $=$ $34,52$
Then look in the → tables for the (volumetric) mass density of the mixture !
$\rho_{12}$ $=$ $0,94587$
Then calculate the volume of the mixture !
$V_{12}$ $=$ $\frac{m_{12}}{rho_{12}}$ $=$ $96,70\;mL$
What do we notice?
The volume of the mixture is not equal to the sum of the volumes of the constituents !
