Answer: The molarity of HCl is 0.041 M.
Explanation:
To calculate the molarity of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is HCl
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is KOH
We are given:
[tex]n_1=1\\M_1=?M\\V_1=20.0mL\\n_2=1\\M_2=0.0300M\\V_2=27.50mL[/tex]
Putting values in above equation, we get:
[tex]1\times M_1\times 20.0=1\times 0.0300\times 27.50\\\\M_1=0.041M[/tex]
Hence, the molarity of HCl is 0.041 M.
