<u>Answer:</u>
<u>For a:</u> The molarity of commercial HCl solution is 12.39 M.
<u>For b:</u> The molality of commercial HCl solution is 16.79 m.
<u>For c:</u> The volume of commercial HCl solution needed is 2.42 L.
<u>Explanation:</u>
We are given:
Mass % of commercial HCl solution = 38 %
This means that 38 grams of HCl is present in 100 grams of solution.
To calculate the volume of solution, we use the equation:
![Density=\frac{Mass}{Volume}](https://tex.z-dn.net/?f=Density%3D%5Cfrac%7BMass%7D%7BVolume%7D)
Density of HCl solution = 1.19 g/mL
Mass of solution = 100 g
Putting values in above equation:
![1.19g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=84.034mL](https://tex.z-dn.net/?f=1.19g%2FmL%3D%5Cfrac%7B100g%7D%7B%5Ctext%7BVolume%20of%20solution%7D%7D%5C%5C%5C%5C%5Ctext%7BVolume%20of%20solution%7D%3D84.034mL)
To calculate the molarity of solution, we use the equation:
![\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20the%20solution%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20solute%7D%5Ctimes%201000%7D%7B%5Ctext%7BMolar%20mass%20of%20solute%7D%5Ctimes%20%5Ctext%7BVolume%20of%20solution%20%28in%20mL%29%7D%7D)
We are given:
Molarity of solution = ?
Molar mass of HCl = 36.5 g/mol
Volume of solution = 84.034 mL
Mass of HCl = 38 g
Putting values in above equation, we get:
![\text{Molality of commercial HCl solution}=\frac{38\times 1000}{36.5\times 84.034}\\\\\text{Molality of commercial HCl solution}=12.39M](https://tex.z-dn.net/?f=%5Ctext%7BMolality%20of%20commercial%20HCl%20solution%7D%3D%5Cfrac%7B38%5Ctimes%201000%7D%7B36.5%5Ctimes%2084.034%7D%5C%5C%5C%5C%5Ctext%7BMolality%20of%20commercial%20HCl%20solution%7D%3D12.39M)
Hence, the molarity of commercial HCl solution is 12.39 M.
To calculate the molality of solution, we use the equation:
![Molarity=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7Bm_%7Bsolute%7D%5Ctimes%201000%7D%7BM_%7Bsolute%7D%5Ctimes%20W_%7Bsolvent%7D%5Ctext%7B%20%28in%20grams%29%7D%7D)
Where,
= Given mass of solute (HCl) = 38 g
= Molar mass of solute (HCl) = 36.5 g/mol
= Mass of solvent = 100 - 38 = 62 g
Putting values in above equation, we get:
![\text{Molality of commercial HCl solution}=\frac{38\times 1000}{36.5\times 62}\\\\\text{Molality of commercial HCl solution}=16.79m](https://tex.z-dn.net/?f=%5Ctext%7BMolality%20of%20commercial%20HCl%20solution%7D%3D%5Cfrac%7B38%5Ctimes%201000%7D%7B36.5%5Ctimes%2062%7D%5C%5C%5C%5C%5Ctext%7BMolality%20of%20commercial%20HCl%20solution%7D%3D16.79m)
Hence, the molality of commercial HCl solution is 16.79 m.
To calculate the molarity of the diluted solution, we use the equation:
![M_1V_1=M_2V_2](https://tex.z-dn.net/?f=M_1V_1%3DM_2V_2)
where,
are the molarity and volume of the concentrated solution
are the molarity and volume of diluted solution
We are given:
![M_1=6M\\V_1=5.00L\\M_2=12.39M\\V_2=?L](https://tex.z-dn.net/?f=M_1%3D6M%5C%5CV_1%3D5.00L%5C%5CM_2%3D12.39M%5C%5CV_2%3D%3FL)
Putting values in above equation, we get:
![6\times 5=12.39\times V_2\\\\V_2=2.42L](https://tex.z-dn.net/?f=6%5Ctimes%205%3D12.39%5Ctimes%20V_2%5C%5C%5C%5CV_2%3D2.42L)
Hence, the volume of commercial HCl solution needed is 2.42 L.