How much concentrated hydrochloric acid solution 36% HCL by mass d=1.18g/ml in milliliters is required to produce 11.0L of 0.296 M HCL?
My calculation
1.18 x 1000ml x 0.36/36.5 = 11.6M
11.6M = 11000ml x 0.296
11.6m = 3256ml
x= 3256ml/11.6
x= 280ml. I am not sure the right answer
How much ethanol, C2H50H, in liters d = 0.789g/ml must be dissolved in water to produce 190.05 of 1.65M C2H50H? I cannot do for your help.
1. I obtained 280.7 which I would round to 281 mL.
2. 190.05 WHAT?
To find the volume of concentrated hydrochloric acid solution needed to produce a certain amount of a desired concentration, you can use the dilution equation:
C1V1 = C2V2
Where:
C1 = initial concentration (36% HCl in this case)
V1 = initial volume (unknown in this case)
C2 = final concentration (0.296 M HCl)
V2 = final volume (11.0 L)
The first step is to convert the initial concentration from a percentage to molarity. To do this, divide the percentage by 100 and multiply by the molar mass of HCl (36.5 g/mol):
Initial concentration (molarity) = (36% HCl / 100) x (36.5 g/mol) = 0.1316 M HCl
Now, we can plug in the values into the dilution equation:
(0.1316 M) x V1 = (0.296 M) x (11.0 L)
To solve for V1, divide both sides of the equation by 0.1316 M:
V1 = (0.296 M x 11.0 L) / 0.1316 M
The units of M (molarity) cancel out, leaving us with the volume in liters:
V1 = (0.296 x 11.0) / 0.1316 L
V1 ≈ 2.488 L
So approximately 2.488 liters of the concentrated hydrochloric acid solution is required to produce 11.0 liters of 0.296 M HCl.
For the second question, to find the volume of ethanol (C2H5OH) needed to produce a certain amount of a desired concentration, you can once again use the dilution equation:
C1V1 = C2V2
Where:
C1 = initial concentration (unknown in this case)
V1 = initial volume (unknown in this case)
C2 = final concentration (1.65 M C2H5OH)
V2 = final volume (190.05 L)
However, the density of ethanol (d = 0.789 g/ml) is given, which means we need to convert the desired final volume from liters to grams. To do this, multiply the volume by the density:
Grams of ethanol = 190.05 L x 0.789 g/ml
Now, we can proceed with the dilution equation:
C1V1 = C2V2
To solve for C1, divide both sides of the equation by V1:
C1 = (C2V2) / V1
Since we have the values for C2, V2, and the grams of ethanol (which can be converted to milliliters), we can substitute them in:
C1 = (1.65 M x 190.05 L) / (190.05 L x 0.789 g/ml)
The units of L (liters) cancel out, and we are left with the concentration in M (molarity):
C1 = 1.65 M / 0.789
C1 ≈ 2.091 M
Now, to find the initial volume of ethanol, V1, we can rearrange the dilution equation:
V1 = (C2V2) / C1
Substituting the values:
V1 = (1.65 M x 190.05 L) / 2.091 M
The units of M (molarity) cancel out, leaving us with the volume in liters:
V1 ≈ 150.714 L
So approximately 150.714 liters of ethanol must be dissolved in water to produce 190.05 liters of 1.65 M C2H5OH.