The volumes of 4N HCl and 10N HCl required to make 1L of 6 N HCl are

solutions mixed are 4N HCl and 10N HCl to form 1l of 6N HClTotal volume of 6N solution = 1lit4N HCl sol.            10N HCl sol.V1= v l                      V2= (1-v) lN1= 4                        N2=10 Resultantnormality ,N=V1N1+V2N2V1+V26=v(4)+(1-v)101 6= 4v+10-10v 6v=4V=23L=0.67L V1=0.67    V2= 1-0.67 = 0.33 L