Greek0000,

The setup shown in the attachment under NOTE inside the box is: 0.10 M C_{3}H_{6}O_{2} x 74 g mol C_{3}H_{6}O_{2} / 1 mol C_{3}H_{6}O_{2}. You should realize that this happens to be a totally wrong setup. If you do what is called dimensional analysis, you can realize it.

First, notice that mol (in 74 g mol) and mol (in 1 mol) will cancel leaving g. M stands for mol / L. So, the final unit left is g mol / L. It is not just g, as it says in the final answer.

Secondly, the correct setup for the conversion factor is: 74 g C_{3}H_{6}O_{2} / 1 mol C_{3}H_{6}O_{2}. This comes from the molar mass. 74 g mol in the numerator is completely wrong.

Multiplying the molarity by molar mass will never give any useful quantity. One needs to know the volume in L. Only then a suitable dimensional analysis will occur. The equation to be used is n = M V and then the resulting n can be converted to g using the molar mass.

This is why Borek refers to it as pure nonsense.