
you want to dilute it 2/.75 times, or 8/3 times. that means one part 2.0M, and 8/31 parts water, or
1 Part originalacid
5/3 part water.
So what is one part? .250/8/3 or .75/8 liter, so one part is .09375 liter of 2M HCl, and .09375*4/3 water, or 0.15625l water.
check: .15625+.09375=.250liter, checks.

how preapare 30.0ml of 0.800m HNO3 FROM A STOCK SOLUTION OF 4.00M HNO3 WHAT VOLUME IN ML OF THE 4.OOM HNO3 SOLUTION WILL YOU NEED?

What will be the volume of 8.52 M HNO3 required to prepare 763 mL of a 1.5 M HNO3 solution?

8.52M X V = 1.5M X 763mL (C1V1 = C2V2)
i.e. Mass of HNO3 in Left = Mass of HNO3 in right
V = (1.5M x 763mL)/ 8.52M = 134.3309mL
Vol of water = 763  134.3309 = 628.6690mL
i.e you add 134.3mL of 8.52M HNO3 to 628.7mL of water.

Describe how to prepare 0.200 L of 0.750 M HCl solution, starting with a 2.00 M HCl solution. What volume (in L) of the 2.00 M solution will you need?