Question
how would ypu prepare 80.00mL of a 7.5% solution of MgSO4 7H2O using solid MgSO4 7H2O and water?
Answers
bobpursley
7.5 percent w/v?
.075(80)= 6 grams MgSO4 = 6/120=.05 moles
but moles of the hydrated salt will have a mass of .05*246.5=12.32 grams.
So add 12.32 grams of the salt, and add water while stirring to the 80ml mark.
Now if the .075 is w/w, it is a different issue.
.075(80)= 6 grams MgSO4 = 6/120=.05 moles
but moles of the hydrated salt will have a mass of .05*246.5=12.32 grams.
So add 12.32 grams of the salt, and add water while stirring to the 80ml mark.
Now if the .075 is w/w, it is a different issue.