1. a) Cm=nd/Vs ⇒ nd=Cm*Vs=0.5*0.2=0.1 moli KOH
M=39+ 16+ 1=56 g/ mol
md=56*0.1=5.6 g KOH
b) ρ=ms/ Vs ⇒ Vs=ms/ ρ=80*1.145=91.6 cm³ sol. acid azotic=0.0916 L
nd=Cm*Vs=4.36*0.0916= 0.399≈0.4 moli HNO₃
M=1+14+3*16=63 g/ mol
md=63*0.4=25.2 g HNO₃
c) ms=ρ*Vs=1.152*400=460.8 g sol. HCl
cp=md*100/ms
md=cp*ms/100=30*460.8/100=138.24 g HCl
2. Cm=nd/Vs
nd=Cm*Vs= 0.02*0.2=0.004 moli HCl
M= 1+35.5=36.5 g/mol
md =0.004*36.5=0.146 g HCl
3. Cm=nd/Vs ⇒ nd=Cm*Vs=0.1*0.5=0.05 moli HNO₃
md=nd*M=0.05*63=3.15 g HNO₃
ms=md*100/cp=3.15*100/20=15.75 s sol. HNO₃
Vs=ms/ρ=15.75/1.119=14.075 ml sol. HNO₃