Posted by Holly on Tuesday, August 30, 2011 at 8:03pm.
Remember M = moles/L (or mmoles/mL).
So we start with how much Cl^-? That is 350 mL x 0.250M = 87.5 mmoles Cl^-. Let y = mL of the MgCl2 to be added.
We have y mL of 0.500 M (in chloride) from MgCl2 to add to it. The total volume will be y mL + 350 mL. Now set that up as mmoles/mL and set it equal to 0.3M.
0.5y mmoles Cl from MgCl2.
350 x 0.250 M = 87.5 mmoles from NaCl.
(0.5y + 87.5)/(y+350) = 0.3M
Solve for y
I estimated it as 87.5 mL of 0.250M MgCl2 (remember it's 0.500 M in Cl^-). You can always check that out. Post your work if you get stuck.
Related Questions
chemistry - Describe how a 250.0 mL aqueous solution of 1.50 M MgCl2 is prepared...
Chemistry - In a clean 250-mL volumetric flask, 125 mL of a 6.0 M hydrochloric ...
chemistry - To make 10 mL of 25 mM sodium standard, what volume of 1 M NaCl ...
chemistry - To make 10 mL of 25 mM sodium standard, what volume of 1 M NaCl ...
chemistry - what is the volume of 0.)250 MgCl2 should be diluted to 250 mL to ...
Chemistry - How many mL of 3.0 M HCL are required to neutralize 250 mL of 2.0 M ...
chemistry - 10.2 g of Pb(NO3)2 are dissolved in 100 mL of water, and added to ...
Chemistry - A buffer is prepared by mixng 150.0 mL of 0.500 M ammonia, 250.0 mL ...
chemistry - What is the pH of the solution that results from mixing the ...
Chemistry - 250-mL volumetric flask, 125 mL of a 6.0 M hydrochloric acid was ...
For Further Reading
