What volumes of 10 M HCl and 3 M HCl should be mixed to get 1L of 6 M HCl solution?

Explanation

Suppose volume of 10 M HCl required to prepare 1 L of 6 M HCl = x litre
Volume of 3 M HCl required = (1 - x) litre Applying molarity equation :
`underset(10 M HCl)(M_(1)V_(1)) + underset(3 M "HCl")(M_(2)V_(2)) = underset(6 M "HCl")(M_(3)V_(3))`
`10 xx x +3(1-x) = 6 xx 1`
`10x + 3-3x=6`
`7x =3` or `x=3/7 = 0.428` L
`therefore` Vol of 10 M HCl required =0.428 L = 428 mL
Vol. of 3 M HCl required `=1-0.428 = 0.572 L`
= 572 mL