Answer : The volume of a 1.50 M HCl solution use should be 0.133 L
Explanation :
Using neutralization law,
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = concentration of HCl solution = 1.50 M
[tex]M_2[/tex] = concentration of another HCl solution = 0.100 M
[tex]V_1[/tex] = volume of HCl solution = ?
[tex]V_2[/tex] = volume of another HCl solution = 2.00 L
Now put all the given values in the above law, we get:
[tex]1.50M\times V_1=0.100M\times 2.00L[/tex]
[tex]V_1=0.133L[/tex]
Therefore, the volume of a 1.50 M HCl solution use should be 0.133 L
