<u>Answer:</u> The mass of HCl present in 500 mL of acid solution is 36.5 grams
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:
![n_1M_1V_1=n_2M_2V_2](https://tex.z-dn.net/?f=n_1M_1V_1%3Dn_2M_2V_2)
where,
are the n-factor, molarity and volume of acid which is HCl
are the n-factor, molarity and volume of base which is NaOH.
We are given:
![n_1=1\\M_1=?M\\V_1=50.00mL\\n_2=1\\M_2=0.500M\\V_2=200.mL](https://tex.z-dn.net/?f=n_1%3D1%5C%5CM_1%3D%3FM%5C%5CV_1%3D50.00mL%5C%5Cn_2%3D1%5C%5CM_2%3D0.500M%5C%5CV_2%3D200.mL)
Putting values in above equation, we get:
![1\times M_1\times 50.00=1\times 0.500\times 200\\\\M_1=\frac{1\times 0.500\times 200}{1\times 50.00}=2M](https://tex.z-dn.net/?f=1%5Ctimes%20M_1%5Ctimes%2050.00%3D1%5Ctimes%200.500%5Ctimes%20200%5C%5C%5C%5CM_1%3D%5Cfrac%7B1%5Ctimes%200.500%5Ctimes%20200%7D%7B1%5Ctimes%2050.00%7D%3D2M)
To calculate the mass of solute, we use the equation used to calculate the molarity of solution:
![\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)
Molar mass of HCl = 36.5 g/mol
Molarity of solution = 2 M
Volume of solution = 500 mL
Putting values in above equation, we get:
![2mol/L=\frac{\text{Mass of solute}\times 1000}{36.5g/mol\times 500}\\\\\text{Mass of solute}=\frac{2\times 36.5\times 500}{1000}=36.5g](https://tex.z-dn.net/?f=2mol%2FL%3D%5Cfrac%7B%5Ctext%7BMass%20of%20solute%7D%5Ctimes%201000%7D%7B36.5g%2Fmol%5Ctimes%20500%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20solute%7D%3D%5Cfrac%7B2%5Ctimes%2036.5%5Ctimes%20500%7D%7B1000%7D%3D36.5g)
Hence, the mass of HCl present in 500 mL of acid solution is 36.5 grams