lt kim sbmptn 2013 kunci

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Pembahasan Latihan Soal Sbmpt n 2013

TKD Saintek - Kim ia

---Cr eated by

ujiantulis.com

( sebagian dar i 5 Paket Latihan Sbmptn)

1. Jaw ab: A

2 HgO (s)



2 Hg (s) + O2(g)

86,4 g 3,2 g = 3 2 32

,

=

0,2 m ol 0,1 m ol

Kadar HgO dalam cuplikan = 0 2 216 g 86 4 g

,



x 100 % = 50 %

2. Jaw ab: B

Kp=Kc(RT)n ,



n = jum lah koefisien gas sebelah kanan – sebelah kiri

Kp=Kc(RT) ,



n = 1

(1) 2 SO3(g)



2 SO2 (g) + O2(g)



n = 3 – 2 = 1

(2) BiCl3(aq) + H2O(l)



BiOCl(s) + 2 HCl(aq)



n = 0 – 0 = 0

(3) PCl5 (g)



PCl3(g) + Cl2(g)



n = 2 – 1 = 1

(4) 2 NO2 (g)



N2O4 (g)



n = 1 – 2 = – 1

3. Jaw ab: E

Nom or atom = jum lah prot on = jum lah elektron = 29

Konfigurasi elekt ron at om 29L : 1s 2

2s2 2p6 3s2 3p6 3d10 4s1

Panit ia SBM PTN Akan M enerapkan 5 Paket Soal Sbmp tn Untuk Ant isipasi perjokian dan kecu rangan (M enurut Dr

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Konfigurasi elektron ion 29L+ : 1s2 2s2 2p6 3s2 3p6 3d10 4s0

[Ar] 3d10

4. Jaw ab: B

Rum us m olekul C4H8O R – C HO : alkanal/ aldehida

Rum us Um um : CnH2nO R – CO – R : alkanon/ ket on

4 3 2 1 3 2 1

R – CHO : CH3 – CH2 – CH2 – CHO dan CH3 – CH – CHO

But anal l

CH3 2 – m etilpropanal

R – CO – R’ : CH3 – CO – CH2 – CH3

But anon

Jum lah isom er = 3

5. Jaw ab: D

Air sadah adalah air yang m engandung ion – ion logam Ca2+ dan M g2+.

a. Air sadah tetap m engandung garam – garam : CaCl2, M gCl2, CaSO4, M gSO4

b. Air sadah sem ent ara m engandung garam – garam : Ca(HCO3)2, M g(HCO3)2 Cara m enghilangkan kesadahan :

a. Air sadah tetap : ditambah soda, Na2CO3 (natrium karbonat)

b. Air sadah sem entara : 1. dipanaskan sam pai m endidih

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6. Jaw ab: A

2CH3COOH + Ca(OH)2



Ca(CH3COO)2 + 2 H2O

50 m L.0,2 M 50 m L. 0,05 M

M ula-m ula : 10 m m ol 2,5 m m ol

Bereaksi : 5 m m ol



2,5 m m ol



2,5 m m ol

Sisa : 5 mmol 0 2,5 mmol

CH3COOH Ca(CH3COO)2 : Penyangga bersifat asam

Ca(CH3COO)2



Ca 2+

+ 2 CH3COO –

2,5 m m ol



5 m m ol

Rumus : [H+] = Ka x 3

3

CH COOH

CH COO









_{= 2,0 x 10}– 5

x 5 m mol/ 100 mL

5 m mol/ 100 mL = 2,0 x 10 – 5

M

pH = 5 – log 2

7. Jaw ab: B

Larutan jenuh L(OH)2 : pH = 9 + 2 log 2 = 9 + log 4

pOH = 5 – log 4



[ OH–] = 4 x 10 – 5 M

L(OH)2 (s)



L2+ (aq) + 2 OH – (aq)

S = 2 x 10 – 5 M



4 x 10 – 5 M

n = 3, Ksp L(OH)2 = 4S3 = 4(2 x 10 – 5)3 = 32 x 10 – 15 = 3,2 x 10 – 14

8. Jaw ab: B

t ½ = 10 tahun, T = lam a m eluruh = 60 t ahun



1 2 60 10 T n t



= 6

Sisa = Ns =

6

1 1 1

64 64 1 gram

2 2 64

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M eluruh = N o – Ns = 64 – 1 = 63 gram

9. Jaw ab: D

KOH + H2SO4 : pH campuran = 2 – log 5 , sisa asam kuat H2SO4 , mol H+ > mol OH–

100 m L.0,1 M 100m L xM [H+] sisa = 5 x 10– 2 M

Volum e cam puran = 200 m L

mmol H+ sisa = 200 x 5 x 10– 2 = 10 mmol

100 m L KOH 0,1 M =10 m m ol



OH– = 10 mmol

100 m L H2SO4 x M = 100 x m m ol



H +

= 200x mmol

mmol H+ sisa = mmol H+ – mmol OH–

= (200 x) – (10) = 200 x – 10 = 10

200 x = 20, x = 0,1 M H2SO4

10. Jaw ab: C 1. CH4 + Cl2

U V

CH3Cl + HCl (reaksi substitusi )

2. C2H5OH 170 oC

C2H4 + H2O (reaksi eliminasi)

3. C2H4 + HCl C2H5Cl (reaksi adisi/ penjenuhan) 4. C2H5OH + CH3COOH CH3COOC2H5 + H2O (reaksi substitusi )

11. Jaw ab: C

Konfigurasi elekt ron ion X 3– : 1s2 2s2 2p6 3s2 3p6 ( m enerim a 3 elekt ron )

Konfigurasi elekt ron at om unsur X : 1s2 2s2 2p6 3s2 3p3



Perioda 3, golongan VA

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M = 10 × % m assa × m assa jenis 10 18 25 1 04

M r 36 5

,





= 5,2 M HCl

13. Jaw ab: C

Isotonik : t ekanan osm ot ik sam a :



1 =



2

M1 R T i1 = M2 RT i2 (RT sam a )

M1 x i1 = M2 x i2

Urea NaOH

6 1000 1 1000 2

60 500 40 200

x



 



(i dari NaOH = 2)

0,2 = 4

x

, x = 0,8 gram NaOH

14. Jaw ab: C

Elekt rolisis larut an NaCl/ C :

NaCl (aq)



Na+(aq) + Cl – (aq)

Katoda (–) : 2 H2O (l) + 2 e



2 OH –

(aq) + H2 (g)

Anoda (+) : 2 Cl–



Cl2(g) + 2e

2 H2O (l) + 2 Cl –



2 OH– (aq) + H2 (g) + Cl2(g)

5,6 L gas H2 (STP) = 5 6

22 4

,

= 0,25 m ol H2

Katoda (–) : 2 H2O (l) + 2 e



2 OH –

(aq) + H2 (g)

0,5 mol e



0,25 mol H2

= 0,5 F

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Sifat oksidat or F2 > Cl2 > Br2 > I2

Cl2 (g) + 2 NaBr (aq)



Br2 (l) + 2 NaCl(aq)