How many ml of concetrated HCl are required to make 2 liters of 3N HCl from concentrated HCl that is 37% w/v and has a specific gravity of 1.2. H=1 Cl=35.5
Dilute about 246.6 ml of HCl 37% into 1000 ml to make 3N HCl
To prepare 2L, dilute 493.2 ml
To answer this question, we need to determine the molarity of the concentrated HCl and then use that information to calculate the volume needed to make the desired concentration.
Step 1: Calculate the molarity of the concentrated HCl.
The formula for molarity (M) is moles of solute divided by liters of solution.
Since we know the concentration of HCl in terms of weight/volume (w/v), we need to convert it to molarity.
First, we need to calculate the moles of HCl in the 37% w/v solution.
To do this, we multiply the weight of the solution by its concentration and divide by the molar mass of HCl:
Weight of 37% HCl solution = 37% * 2000 ml * 1.2 g/ml = 888 g
Molar mass of HCl = (1 g/mol * 1) + (35.5 g/mol * 1) = 36.5 g/mol
Moles of HCl = 888 g / 36.5 g/mol = 24.3 mol
Next, we need to calculate the volume of 3N HCl needed.
The formula to convert from normality (N) to molarity (M) is Molarity = Normality * Equivalent weight.
In this case, the equivalent weight of HCl is equal to its molar mass since it's a monoprotic acid.
Equivalent weight of HCl = Molar mass of HCl = 36.5 g/mol
To determine the molarity (M) of 3N HCl, we rearrange the formula as follows:
Molarity = Normality / Equivalent weight
Molarity = 3N / 36.5 g/mol = 0.082 M
Step 2: Calculate the volume of concentrated HCl needed.
Using the formula M1V1 = M2V2, where M1 is the molarity of the concentrated HCl, V1 is the volume of the concentrated HCl, M2 is the desired molarity (0.082 M), and V2 is the desired volume (2 liters), we can solve for V1.
(0.082 M)(V1) = (24.3 mol)(2 L)
V1 = (24.3 mol * 2 L) / 0.082 M
V1 = 593.9 ml
Therefore, approximately 594 ml of concentrated HCl is required to make 2 liters of 3N HCl.