Розв’язування

Äàíî:

m_1ð–íó = 400 ã

₁ = 12 %

V_2ð–íó=200 ìë

Ñì = 0,2 Ì

₂ = 1,1 ã/ìë

Ì(ÊÎÍ)= 56ã/ìîëü

₃ –?

m_p.p.100 m_ð–íó 12400

=  ;  m_1KOH =  =  = 48 (ã)

m_ð–íó 100 100

m_KOH

Ñì =   m_2KOH = ÑìÌ_KOHV_ð–íó =560,20,2=

Ì_KOHV_ð–íó

= 2,24 (ã); m₃(ÊÎÍ) = 48 + 2,24 = 50,24 (ã)

m_2ð–íó = V_2ð–íó_2ð–íó = 2001,1 = 220 (ã)

m_3ð–íó = 400 + 220 = 620 (ã)

50,24_.100

₃ =  = 8,1 (%)

620

³äïîâ³äü: 8,1 (%)

Äàíî:

 = 36,2%

 = 1,18 ã/ìë

V_ð–íó = 1 ë

Ñì –?

Молярн³сть – це к³льк³сть молей розчинено¿ речовини в 1 л розчину. m_ð–íó = V_ð–íó_ð–íó = 10001,18 = 1180 (ã)

m_p.p.100 _.m_ð–íó 36,21180

=  ;  m_HCl =  =  = 427,16(ã)

m_ð–íó 100 100

m_HCl 427,16

Ñì =  =  = 11,7 (ìîëü/ë)

Ì_HClV_ð–íó 36,51

³äïîâ³äü: 11,7 ìîëü/ë

Äàíî:

m(Í₂Î)=1000ã

m(ÊÎÍ)=224 ã

 = 1,23 ã/ìë

Ì(ÊÎÍ) =

= 56 ã/ìîëü

 –? Ñì –?

C_m –?

n(Í₂Î) –?

n(KOH) –?

m_ð–íó­ = m(Í₂Î) + m(ÊÎÍ) = 1000 + 224 = 1224 (ã)

V_ð–íó = m_ð–íó /_ð–íó = 1224/1,23=995,12 (ìë)=0,995ë

m_p.p.100 224_.100

=  =  = 18,3 (%);

m_ð–íó 1224

m_KOH 224

Ñì =  =  = 4,02 (ìîëü/ë)

Ì_KOHV_ð–íó 560,995

m_KOH1000 2241000

Ñ_m =  =  = 4,0 (ìîëü/êã)

Ì_KOHm_ð–êà 561000

_KOH 224/56

n_KOH =  =  = 0,067

_KOH + í₂Î 224/56 + 1000/18

í₂Î 1000/18

ní₂Î =  =  = 0,933

_KOH + í₂Î 224/56 + 1000/18

³äïîâ³äü: 18,3 %; 4,02 ìîëü/ë; 4,0 ìîëü/êã; 0,933; 0,067

Äàíî:

m(C₆H₅NO₂) = 1 ã

m(C₆H₆) = 10 ã

Å = 2,57⁰Ñ

Ê = 5,1⁰Ñ

t_êèï.(C₆H₆)=80,2⁰Ñ

t_{замерз.}(C₆H₆)=5,4⁰Ñ

t_{êèï. ð–íó} –?

t_{замерз. р–ну} –?

Ì(C₆H₅NO₂) =

= 123,11 ã/ìîëü

Зг³дно другого закону Рауля

Åm_ð.ð.1000 2,571_.1000

t_êèï. =  =  = 2,09⁰C;

Ì_ð.ð.m_ð–êà 123,11_.10

Km_ð.ð.1000 5,11_.1000

t_{замерз.} =  =  = 4,14⁰Ñ;

Ì_ð.ð.m_ð–êà 123,11.10

t_{êèï. ð–íó}=t_êèï.(C₆H₆)+t_êèï.=80,2+ 2,09=82,29⁰C.

t_{замерз. р–ну} = t_{замерз.}(C₆H₆) – t_{замерз.}=

= 5,4 – 4,14 = 1,26⁰C

³äïîâ³äü: 82,29⁰C; 1,26⁰C

Äàíî:

m_ð.ð. = 2,5 ã

V_ð–íó= 5 ë

Ò = 20+273=293 Ê

 = 23 êÏà

Ì –?

Зг³дно закону Вант–Гоффа  = С_ÌRT,

m_p.p. m_p.p. RT m_p.p. RT

Ñì = ;  =   Ì_p.p.=  =

Ì_p.p.V_ð–íó Ì_p.p.V_ð–íó V_ð–íó

2,58,314293

_.=  = 52,96 (ã/ìîëü)

235

³äïîâ³äü: 52,96 ã/ìîëü

Äàíî:

Ð₀ =3,166 êÏà

 = 5%

Ð₁ –?

Зг³дно ² закону Рауля Р₁ = Ð₀n(H₂O), äå n(H₂O) – мольна частка води.

í₂Î 95/18

ní₂Î =  =  = 0,984

CO(NH₂)₂ + í₂Î 5/60 + 95/18

P₁ = 3,1660,984 = 3,117 (êÏà)

³äïîâ³äü: 3,117 êÏà