

InterviewSolution
Saved Bookmarks
1. |
What is the normality of a solution that results from mixing 7.4 g of `Ca(OH)_2, 500` " mL of " `1 M HNO_3` and `10.0 mL` of `H_2SO_4` (specific gravity`=1.2,49%H_2SO_4` by weight)? The total volume of the solution was made to 1L after adding water? |
Answer» (a). `Mw[Ca(OH)_2]=40+34=74` `7.4g. Of Ca (OH)_2=(7.4)/(74)-=0.1mol=10m` moles `=10xx2mEq` (n-factor`=2)` `-=20mEq` (b). `500 " mL of " 1 M HNO_3=500xx1xx1` (n-factor) `-=20mEq` (b). `500 " mL of " 1 M HNO_3=00xx1xx1` (n-factor) `-=500 mEq` (c). Normality of `H_2SO_4` `=(Wxx1000)/(Ewxx"Volume of solution in mL")` `{:[(Mw=98,"n-factor=2),(Ew=(98)/(2)=49g)]:}` or Normality of `H_2SO_4` `=(%"by weight"xx"density"("or specific gravity"))/(Ew)` `=(49xx10xx1.2)/(49)=12N` `m" Eq of "H_2SO_4=NxxV=12xx10=120 mEq` (d). Total m" Eq of "acid `=500+120=620 mEq` (e). m" Eq of "`Ca(OH_2)=20mEq` (f). m" Eq of "acid left `=620-20=600 mEq` (g). Normality of solution `=(mEq)/(V(mL))=(600)/(1000)=0.6N` |
|