1.) How much ethanol, C2H5OH, in liters (d=0.789g/ml) must be dissolved in water to produce 190.5 of 1.65 M C2H5OH? (Molarity problem.)

2.) How much concentrated hydrochloric acid solution (36.0% HCl by mass, d = 1.18g/mL}), in milliliters, is required to produce 11.0 L of 0.296 M HCl?

1.) How much ethanol, C2H5OH, in liters (d=0.789g/ml) must be dissolved in water to produce 190.5 of 1.65 M C2H5OH? (Molarity problem.)


190.5 WHAT for crying out loud?barrels, freight cars, lakes, I will assume you meant mL. If not that, adjust values below to what you should have written.
M = moles/L and moles = grams/molar mass.

1.65M = moles/0.1905. Calculate moles.
moles = gram/molar mass.
You know moles and molar mass, calculate grams.
From the density, calculate volume of ethanol to use and convert to liters.

molarity=mass/molmass*volumeinLiters

1) massEthanol=density*volumeEthanol

so volumeEthanol=molarity*molmass*Volumesolution/density

Watch units.

2)work out the similar steps for the HCL.

2.) How much concentrated hydrochloric acid solution (36.0% HCl by mass, d = 1.18g/mL}), in milliliters, is required to produce 11.0 L of 0.296 M HCl?


First calculate the molarity of the concentrated HCl.
1.18 g/mL x 1000 mL - mass of 1 L = 1180 g.
How much of that is HCl? 36% so,
1180 x 0.36 = 424.8 g HCl.
How many moles is that?
424.8/36.5 = 11.6 M
Now use the dilution formula to determine what you want to prepare.
mL x M = mL x M
Check my work. I estimated the molar masses and rounded here and there so you need to go through the problem yourself and do those numbers right.

The second one is correct. Regarding the first one though, I calculated the grams to 6.8 g of C2H5OH. I don't quite understand the next step (regarding the density, etc)

that was me who posted the comment above. my friend had the same question.

I don't know how you obtained 6.8 g.

The part about the density:
The problem asks for VOLUME of ethanol to add to make ?? soln of ??M.
mass = volume x density
You calculate mass from above, substitute density here of 0.789, and solve for volume (in mL). Then convert to L. Look at your math. I don't get 6.18 grams (more like 15 g or so--in round numbers).

To solve these types of problems, we need to use the formula for molarity:

Molarity (M) = moles of solute / liters of solution

Let's break down each problem:

1.) How much ethanol, C2H5OH, in liters (d=0.789g/ml) must be dissolved in water to produce 190.5 of 1.65 M C2H5OH? (Molarity problem.)

To find the answer, we need to follow these steps:

Step 1: Convert the given volume from mL to liters.
190.5 mL = 190.5 / 1000 = 0.1905 L

Step 2: Calculate the number of moles of C2H5OH using the molarity formula.
Molarity (M) = moles of C2H5OH / liters of solution

Rearranging the formula, we have:
moles of C2H5OH = Molarity (M) x liters of solution

moles of C2H5OH = 1.65 M x 0.1905 L = 0.314325 moles

Step 3: Convert moles of C2H5OH to grams using the molar mass of C2H5OH.
The molar mass of C2H5OH (ethanol) is:
C = 12.01 g/mol
H = 1.01 g/mol (there are 6 hydrogens)
O = 16.00 g/mol
Summing up these values, we get:
12.01 + (1.01 x 6) + 16.00 = 46.07 g/mol

mass of C2H5OH = moles of C2H5OH x molar mass of C2H5OH
mass of C2H5OH = 0.314325 moles x 46.07 g/mol = 14.47 g

Step 4: Convert grams of C2H5OH to volume using density
We are given the density of ethanol, which is:
density (d) = mass / volume

Rearranging the formula, we have:
volume = mass / density

volume = 14.47 g / 0.789 g/mL = 18.35 mL

Converting mL to liters:
volume = 18.35 mL / 1000 = 0.01835 L

Therefore, to produce 190.5 mL of 1.65 M C2H5OH, we need to dissolve approximately 0.01835 L (or 18.35 mL) of ethanol in water.

2.) How much concentrated hydrochloric acid solution (36.0% HCl by mass, d = 1.18g/mL}), in milliliters, is required to produce 11.0 L of 0.296 M HCl?

To find the answer, follow these steps:

Step 1: Calculate the moles of HCl required using the molarity formula.
Molarity (M) = moles of HCl / liters of solution

Rearranging the formula, we have:
moles of HCl = Molarity (M) x liters of solution

moles of HCl = 0.296 M x 11.0 L = 3.256 moles

Step 2: Calculate the mass of HCl required using the mass percent formula.
mass of HCl = mass percent of HCl x mass of solution

mass percent of HCl = 36.0% = 0.36
mass of solution = volume of solution x density
volume of solution = 11.0 L = 11.0 x 1000 = 11000 mL

mass of solution = 11000 mL x 1.18 g/mL = 12980 g
mass of HCl = 0.36 x 12980 g = 4672.8 g

Step 3: Convert grams of HCl to volume using density.
density (d) = mass / volume

Rearranging the formula, we have:
volume = mass / density

volume = 4672.8 g / (1.18 g/mL) = 3959.32 mL

Therefore, approximately 3959.32 mL (or 3.95932 L) of concentrated hydrochloric acid solution is required to produce 11.0 L of 0.296 M HCl.