4% w/w MgCl2 = 4.00 g MgCl2/100 g soln.
volume soln = mass/density = 100g/1.15 g/mL = 86.96 mL.
moles MgCl2 in 4.00 g = 4.00/95.21 = 0.04201 moles.
M Mg^+2 = moles Mg^+2/L soln
M Cl^- = moles Cl^-/L soln.
what are the molarities of Mb^2+ ions and Cl- ions in an aqueous solution of 4.00% by the mass magnesium chloride, with a density of 1.15g mL-1?
2 answers
0.04201/0.08696 = 0.483 M of [Mg^2+]
but how do I get the [Cl-]? the answer is suppose to be 0.966M
but how do I get the [Cl-]? the answer is suppose to be 0.966M