Question

How many grams of methanol (CH3OH, FM 32.04) arecontained in0.100 L of 1.71 M aqueous methanol (i.e., 1.71 molCH3OH/L solution)?

Short answer

The mass of the methanol is 5.48 g.

Step-by-step solution

Finding the mass of methanol

The mass of methanol can be calculated by using the formula of Molarity.

Given values

The volume= 0.100 L

The molar mass of CH3OH= 32.04 g/mol

The molarity of CH3OH= 1.71 M

Find the value of mass

Use the formula below:

$\mathrm{Molarity}=\frac{\mathrm{No}.\mathrm{of}\mathrm{moles}}{\mathrm{Volume}\mathrm{in}\mathrm{L}}$

Whereas, the no. of moles=$\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Molar}\mathrm{mass}}$

Replace the given values:

$1.71\mathrm{M}=\frac{\mathrm{Given}\mathrm{mass}}{{\displaystyle \frac{32.04}{1.100\mathrm{L}}}}$

Given mass role="math" localid="1663307729434" $=1.71\times 32.04\times 0.100=5.48\mathrm{g}$

Thus, the mass of methanol is 5.48 g.