Question

What is the molarity of a solution that contains 125g CH3OH in 0.25L of solution

Answer 1

Well
\[Molarity=\frac{ moles of solute }{ Liters of Solution }\]
Since we were only given the amount of CH3OH, we will have to add up the atomic masses of the elements in it.
12+3+16+1=32g of CH3OH per mole.
Then we'd do factor label to convert it to moles.
\[\frac{ 125g }{ 1 } \times \frac{ 1mol }{ 32g } = \frac{ 125 }{ 32 } moles\]
125g of CH3OH is 3.90625 moles. So we would put that into the equation above.
\[\frac{ 3.90625mol }{ .25L } = 15.625\]
But in this answer we are limited to three significant figures due to the fact that the original values given all had 3 significant figures.
So your final answer would be 15.6M.