4/.
c% = mdx100/ms
=> md.CaCl2 = msxc%/100
= 44,4x10/100 = 4,44 g CaCl2
4,44 g md g
CaCl2 + 2AgNO3 --> Ca(NO3)2 + 2AgCl
111 2x170
=> md.AgNO3 = 13,6 g
din c% => ms = mdx100/c%
= 13,6x100/20 = 68 g sol. AgNO3
5/.
x mL sol H2SO4
stim ca Cm = roxc%x10/miu
=> c% = miuCm/10ro
= 98x9/10x1,47 = 60%
aplicam regula dreptunghiului:
c1=60% --------------------> 48-30 = 18 parti masice sol. c1%
c.f.=48%
c2=30% -------------------> 60-48 = 12 parti masice sol. c2%
---------------------------------------------------------
total parti masice = 18+12 = 30
30 parti ............. 18 parti sol. c1% ............... 12 parti sol. c2%
ms.tot ............... ms1 .................................... 200 g
=> ms.tot = 30x200/12 = 500 g sol
=> ms1 = 18x200/12 = 300 g sol.
stim ca ro = ms/Vs = ms/x
=> x = ms1/ro = 300/1,47 = 204,08 mL sol. H2SO4
