Molarity problem:
assuming the density of vinegar is 1.0 g/ml, what is the molarity of vinegar? (use the percent by mass of acetic acid in vinegar from question 1 which is 5.3%). the molar mass of acetic acid is 60.05 g/mol.

Density = 1.0 g/mL

5.3 (w/w)% => 5.3 / 100 => 0.053

molar mass acetic acid = 60.05 g/mol

M = 1000.  w . D / molar mass

M = 1000 . 0.053 . 1.0 / 60.05

M = 53 / 60.05

M = 0.882 mol/L

hope this helps!

There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution. 

5.3 % = 5.3 g acetic acid / 100 g solution

Molarity = 5.3 g acetic acid / 100 g solution ( 1.0 g solution / .001 L solution ) ( 1 mol acetic acid/ 60.05 g acetic acid ) = 0.8826 mol / L solution

Organisms are more likely to be more preserved in mud than sand because mud is moist and wet, while sand is very dry.

Ok where is the image ?