How much methanol, CH3OH (d = 0.792g/mL), in milliliters, must be dissolved in water to produce 2.15L of 0.470 M CH3OH?
To find out the amount of methanol (CH3OH) in milliliters required to produce a specific concentration in a specific volume, we need to use the equation:
\(M_1V_1 = M_2V_2\)
Where:
M1 = initial concentration of the methanol solution
V1 = initial volume of the methanol solution
M2 = final concentration of the methanol solution
V2 = final volume of the methanol solution
Given:
M1 = Unknown (initial concentration)
V1 = Unknown (initial volume)
M2 = 0.470 M (final concentration)
V2 = 2.15 L (final volume)
We can rearrange the equation to solve for V1:
\(V_1 = \frac{{M_2 \cdot V_2}}{{M_1}}\)
To proceed with the calculations, we need to convert the final volume from liters to milliliters:
\(V_2 = 2.15 \text{ L} \times 1000 \text{ mL/L}\)
Let's substitute the given values into the equation to calculate the initial volume V1:
To solve this problem, we need to use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
First, let's calculate the number of moles of CH3OH needed using its molarity and volume:
moles of CH3OH = Molarity × volume of solution (in liters)
= 0.470 M × 2.15 L
= 1.0105 moles
Next, we'll convert this number of moles to grams using the molecular weight of CH3OH:
molecular weight of CH3OH = 12.01 g/mol (C) + 1.01 g/mol (H) + 16.00 g/mol (O) + 1.01 g/mol (H)
= 32.04 g/mol
grams of CH3OH = moles of CH3OH × molecular weight of CH3OH
= 1.0105 moles × 32.04 g/mol
≈ 32.35 g (rounded to two decimal places)
Now, we have the mass of CH3OH required. Since we are given the density of CH3OH, we can then calculate its volume using the formula:
volume (in milliliters) = mass (in grams) / density (in g/mL)
= 32.35 g / 0.792 g/mL
≈ 40.92 mL (rounded to two decimal places)
Therefore, approximately 40.92 mL of methanol must be dissolved in water to produce 2.15 L of 0.470 M CH3OH.