What is the molarity of #"19.00 cm"^3"# of glacial acetic acid at #"25"^@"C"#?

Question

Noah Aldrich · Accepted Answer

C=1.146 M

Your solution will have a molarity of 1.146 M. To solve this problem, you need the density of glacial acetic acid, also known as anhydrous acetic acid. You must ascertain the mass of dissolved glacial acetic acid because molarity is calculated by dividing moles of solute—in this case, glacial acetic acid—by liters of solution. Glacial acetic acid has a density of "1.05 g/mL" (https://tutor.hix.ai), meaning that dissolving a large amount of #19.00cancel("mL") * "1.05 g"/(1cancel("mL")) = "19.95 g acetic acid"# NOTE: Since one cubic centimeter is equal to one milliliter, I won't bother doing an actual conversion when I use milliliters (mL) for volume. Now find the number of moles by using the molar mass of acetic acid. #19.95cancel("g") * "1 mole"/(60.05cancel("g")) = "0.33222 moles"# Lastly, calculate the molarity of the solution using its volume. #C = n/V = "0.33222 moles"/(290.0 * 10^(-3)"L") = "1.14559 M"# The response, rounded to four sig figs, is #C = color(green)("1.146 M")#

Peyton Adams · Answer

1.1445 mol/L, or 1.1445 M, is the molarity. Because you were not given the molarity of the glacial acetic acid, you will have to calculate it using its volume, density, and molar mass. This is why your teacher gave you the temperature (density changes with temperature). The density will have to come from a resource. The molecular formula for glacial acetic acid is #"CH"_3"COOH"#, so you can calculate its molar mass, but I looked it up.  A substance's molarity is measured in moles of solute per liter of solution; molarity is represented by the symbol M and is expressed in units of mol/L. Glacial acetic acid will be referred to as GA from now on. GIVEN/KNOWN:
#V_1=19.00"cm^3#
#"density of GA at 25"^("o")"C"=1.049"g/cm"^3#
https://tutor.hix.ai
#"molar mass of GA"="60.05g/mol"#
https://tutor.hix.ai
#V_2=290.0 "cm"^3"=290.0mL=0.2900L# UNKNOWN:
#M_"GA"# RESOLUTION: Mass of GA.
Determine the mass of GA using its given volume and its density.
mass = density x volume = #(1.049 g)/(1 cancel(cm^3))xx19.00 cancel(cm^3)# = #"19.931 g"#
mass of GA = #"19.931 g"# Moles of GA 
Determine the moles of GA by using its mass and molar mass of 60.05 g/mol.
#19.931 cancel("g")xx(1 "mol")/(60.05 cancel("g")# = #"0.33191 mol GA"# Moles of GA = #"0.33191 mol"# GA's molarity can be calculated by dividing its molecular weight (mol GA) by the number of liters in the solution. #M = "mol/L"#= #"0.33191 mol"/"0.2900 L"= 1.1445 "mol/L"#

Ellie Andrews · Answer

undefined To calculate the molarity of a solution, you need to know the volume of the solution in liters and the number of moles of solute dissolved in that volume. Since glacial acetic acid is the solute, you also need its molar mass. With this information, you can use the formula:

$$ \text{Molarity} (M) = \frac{\text{Number of moles of solute}}{\text{Volume of solution (in liters)}} $$

First, convert the volume of glacial acetic acid from cubic centimeters (cm^3) to liters:

$$ \text{Volume} = 19.00 \, \text{cm}^3 \times \frac{1 \, \text{mL}}{1 \, \text{cm}^3} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} $$
$$ \text{Volume} = 0.019 \, \text{L} $$

Next, you need to know the molar mass of glacial acetic acid (CH3COOH), which is approximately 60.05 g/mol.

Now, we can calculate the number of moles of glacial acetic acid using the volume and the molarity formula:

$$ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} $$

However, since the density of glacial acetic acid is close to that of water, we can assume that 19.00 cm^3 of glacial acetic acid weighs approximately 19.00 g.

$$ \text{Number of moles} = \frac{19.00 \, \text{g}}{60.05 \, \text{g/mol}} $$

$$ \text{Number of moles} = 0.316 \, \text{mol} $$

Finally, plug the number of moles and the volume into the molarity formula:

$$ \text{Molarity} = \frac{0.316 \, \text{mol}}{0.019 \, \text{L}} $$

$$ \text{Molarity} = 16.63 \, \text{M} $$

Therefore, the molarity of 19.00 cm^3 of glacial acetic acid at 25°C is approximately 16.63 M.