how many molecules of liquid ethanol (CH3H2OH) are in 1.00 mL of ethanol, given its density is 0.789 g/mL?

Question

rogue · Answer

Step 1: Get mass of the 1.00 mL ethanol
Density = mass/Volume$$D = \frac {m}{V} \rightarrow m = DV$$$$m = 0.789 g/mL \times 1 mL = 0.789 g$$
Step 2: Get grams of ethanol to moles of ethanol. There are 46.07 grams of ethanol per mole of ethanol.$$0.789 g \times \frac {1 mol}{46.07 g} = 0.0171 mol$$
Step 3: Get moles ethanol to molecules of ethanol. There are 6.022 x 10^23 molecules of ethanol per mole ethanol.$$0.0171 mol \times \frac {6.022 \times 10^{23} molecules}{1 mol}  = 1.03 \times 10^{22} molecules$$