REMIDI KIMIA KLAS XII SMA SHAFTA SBY
Untuk nomor absen ganjil
1.Sebanyak 18 gram glukosa (Mr= 180) dilarutkan dalam 72 gram air. Pada suhu
tertentu tekanan uap air 20,1 cm Hg. Hitunglah penurunan tekanan uap larutan
glukosa tersebut!
Untuk nomor absen genap
2.Ke dalam 600gram air dilarutkan 27 gram senyawa non elektrolit. Larutan itu
mendidih pada suhu 100,13 derajat C. Hitunglah Mr senyawa tersebut !
Jawaban paling lambat tanggal 31 jan 2008 jam 9 pagi
posted by dwi agus dh at
7:05 PM
28 Comments:
pak saya irwan (XII-IPA) SMA SHAFTA. q kerjain remidinya di e-mail pak agus (dwiagus11@yahoo.com) aja ya pak coz klo ngerjain lewat comment sulit, jadinya gak kebaca
1x lagi maaf ya pak?
January 24, 2008 at 7:39 AM
pak katanya bimbelnya yang ngajar bapak? tyuz knp kmrn yang ngajar tu guru laen?????
January 24, 2008 at 8:30 AM
Nama : Achmad Sholekhudin
Kelas : XII-IPA / 05
Diket : w glukosa = 18 gr
w air = 72 gr
Mr C6H12O6 = 180
Mr H2O = 18
P° = 20,1mmHg
Dit : Delta P………………...?
Jawab :
n1 C6H12O6 = 18 gr x 1mol/180 gr = 0,1 mol
n2 H2O = 72 gr x 1mol/18 gr = 4 mol
X C6H12O6 = n1 / n1+n2
= 0,1 / 0,1+4 = 0,024
Delta P = P° x X C6H12O6
= 2010 mmHg x 0,024
= 48,24 mmHg
January 25, 2008 at 1:17 AM
pak sorry, Q nulis Lanbang deltanya Pake Tulisan biasa.
coz...
dalam Comment lambangnya ga' kebaCa.....
ya udah Gtu aja se kali Lg SoRy ya Pak..
Achmad Sholekhudin Brownson
XII-IPA / 05
January 25, 2008 at 1:29 AM
pak mang bener ta bapak mirip SOGI (extravaganza)? tapi kalo tak lihat2 c mirip juga, tyuz tak lihat2 lagi kok tambah semakin mirip ya? bukan cuma aku lho pak anak XII-IPA juga ngira bapak mirip SOGI, tapi tenag pak masih gantengan bapak kok.
jangan marah ya pak i'm just laught ok....................
January 25, 2008 at 1:36 AM
pak agus saya perwakilan dari kelas XII IPA mohon doanya ya pak agar kami sekelas bisa lulus ujian nasional
terutama beberapa murid yang bapak anggap sebagai calon murid yang tidak lulus
terima kasih ya pak
saya akan berusaha selalu dengan belajar dan berdoa
January 26, 2008 at 10:24 PM
yusuf ferYANTO
XII IPA
Diket : massa glukosa = 18 gr
massa air = 72 gr
Mr C6H12O6 = 180
Mr H2O = 18
P° = 20,1cmHg =201 mmHg
Dit : Delta P………………...?
Jawab :
n1 C6H12O6 = 18 gr x 1mol/180 gr = 0,1 mol
n2 H2O = 72 gr x 1mol/18 gr = 4 mol
X C6H12O6 = n1 : (n1+n2)
= 0,1 : (0,1+4) = 0,024
Delta P = P° x X C6H12O6
= 201 mmHg x 0,024
= 4,90 mmHg
thanks doakan agar kelas XII IPA lulus semua untuk sma shafta
mari pak belajar yang giat bwt unas
January 27, 2008 at 6:26 PM
dik : w1= 600 gr
w2= 27 gr
Tb= 100,13 derajat C
Kb air= 0,51 derajat C/molal
dit : Mr senyawa tsb ?
jwb : delta Tb= 100,13 - 100 = 0,13
delta Tb= w2/Mr.1000/w1.Kb
0,13= (27/Mr.1000/600).0,51
0,13= 13770/600Mr
78 Mr= 13770
Mr = 13770/78
Mr = 176,53
January 27, 2008 at 9:02 PM
nama : rizal aditya
kelas i2 ipa
hal : her kimia
diketahui : w1= 600 gr
w2= 27 gr
Tb= 100,13 derajat
Kb air= 0,51 derajat C/molal
dit : Mr senyawa tsb ?
jwb : delta Tb= 100,13 - 100 = 0,13
delta Tb= w2/Mr.1000/w1.Kb
0,13= (27/Mr.1000/600).0,51
0,13= 13770/600Mr
78 Mr= 13770
Mr = 13770/78
Mr = 176,6
January 28, 2008 at 6:05 AM
Nama : M. Rizal K.
Kelas : XII-IPA1
No : 20
Diket : W1 = 600 gram
W2= 27 gram
Tb = 100,13°C
Dit : Mr…………………?
Jawab :
Delta Tb = (100,13-100)°C = 0,03°C
Delta Tb = W2 / Mr x 1000 / W2 x Kb
0,13 = 27/Mrx1000/600x0,51°C/molal
0,13 mr = 27000 / 600 x 0,51°C
= 0,0176 gr/mol
January 28, 2008 at 9:25 PM
Nama : Hilmi Sanjaya.
Kelas : XII-IPA1
No : 12
Diket : W1 = 600 gram
W2= 27 gram
Tb = 100,13°C
Dit : Mr…………………?
Jawab :
Delta Tb = (100,13-100)°C = 0,03°C
Delta Tb = W2 / Mr x 1000 / W2 x Kb
0,13 = 27/Mrx1000/600x0,51°C/molal
0,13 mr = 27000 / 600 x 0,51°C
= 0,0176 gr/mol
January 28, 2008 at 9:32 PM
Nama : Irwan Rudiansyah.
Kelas : XII-IPA1
No : 20
Diket : W1 = 600 gram
W2= 27 gram
Tb = 100,13°C
Dit : Mr…………………?
Jawab :
Delta Tb = (100,13-100)°C = 0,03°C
Delta Tb = W2 / Mr x 1000 / W2 x Kb
0,13 = 27/Mrx1000/600x0,51°C/molal
0,13 mr = 27000 / 600 x 0,51°C
= 0,0176 gr/mol
pak q nggarap lagi cz yg kmrn salah
sry ya pak
January 28, 2008 at 9:34 PM
nama; setyo lukman
kelas; XII Ipa
Dik: Mp: 72 gram
Mrp: 18
Po: 20,1 cm Hg
Mt: 18 gram
Mrt: 180
Dit: DP
Jwb:
Mol t = gram = 18:180 = 0.1 mol
Mr
Mol p = Gram = 72 :18 = 4mol
Mr
DP = Po . Xt
= Po nt
nt + np
= 20.1 0.1
0.1 + 4
= 20.1 . 0.024
=0.482 cm Hg
January 29, 2008 at 12:38 AM
Nama : m.alif arifin
kelas XII IPA
absen: 22
hal : remidi kimia
Diket : W1 = 600 gram
W2= 27 gram
Tb = 100,13°C
Dit : Mr…………………?
Jawab :
Delta Tb = (100,13-100)°C = 0,03°C
Delta Tb = W2 / Mr x 1000 / W2 x Kb
0,13 = 27/Mrx1000/600x0,51°C/molal
0,13 mr = 27000 / 600 x 0,51°C
= 0,0176 gr/mol
January 29, 2008 at 3:32 AM
Nama : AHMAD HAFID
kelas XII IPA
absen: 2
hal : REMIDI KIMIA
Diket : W1 = 600 gram
W2= 27 gram
Tb = 100,13°C
Dit : Mr…………………?
Jawab :
Delta Tb = (100,13-100)°C = 0,03°C
Delta Tb = W2 / Mr x 1000 / W2 x Kb
0,13 = 27/Mrx1000/600x0,51°C/molal
0,13 mr = 27000 / 600 x 0,51°C
= 0,0176 gr/mol
January 29, 2008 at 3:35 AM
pak saya alif saya sudah ngerjain remidi kimia
kalau salah mohon di maafkan......
makasih wassalamualaikum wr. wb.
January 29, 2008 at 3:39 AM
Nama : Luqman Idham Arzha
Kelas : XII-IPA / 19
Diket : w glukosa = 18 gr
w air = 72 gr
Mr C6H12O6 = 180
Mr H2O = 18
P° = 20,1mmHg
Dit : Delta P………………...?
Jawab :
n1 C6H12O6 = 18 gr x 1mol/180 gr = 0,1 mol
n2 H2O = 72 gr x 1mol/18 gr = 4 mol
X C6H12O6 = n1 / n1+n2
= 0,1 / 0,1+4 = 0,024
Delta P = P° x X C6H12O6
= 2010 mmHg x 0,024
= 48,24 mmHg
January 29, 2008 at 7:48 PM
Nama : Suryo N. Effendy
kelas XII IPA
absen: 38
hal : remidi kimia
Diket : W1 = 600 gram
W2= 27 gram
Tb = 100,13°C
Dit : Mr…………………?
Jawab :
Delta Tb = (100,13-100)°C = 0,03°C
Delta Tb = W2 / Mr x 1000 / W2 x Kb
0,13 = 27/Mrx1000/600x0,51°C/molal
0,13 mr = 27000 / 600 x 0,51°C
= 0,0176 gr/mol
January 29, 2008 at 7:53 PM
Nama : Randhie Jihadillah
Kelas : XII-IPA / 47
Diket : w glukosa = 18 gr
w air = 72 gr
Mr C6H12O6 = 180
Mr H2O = 18
P° = 20,1mmHg
Dit : Delta P………………...?
Jawab :
n1 C6H12O6 = 18 gr x 1mol/180 gr = 0,1 mol
n2 H2O = 72 gr x 1mol/18 gr = 4 mol
X C6H12O6 = n1 / n1+n2
= 0,1 / 0,1+4 = 0,024
Delta P = P° x X C6H12O6
= 2010 mmHg x 0,024
= 48,24 mmHg
January 29, 2008 at 7:55 PM
Nama : M, Abdul Q.
Kelas : XII-IPA / 23
Diket : w glukosa = 18 gr
w air = 72 gr
Mr C6H12O6 = 180
Mr H2O = 18
P° = 20,1mmHg
Dit : Delta P………………...?
Jawab :
n1 C6H12O6 = 18 gr x 1mol/180 gr = 0,1 mol
n2 H2O = 72 gr x 1mol/18 gr = 4 mol
X C6H12O6 = n1 / n1+n2
= 0,1 / 0,1+4 = 0,024
Delta P = P° x X C6H12O6
= 2010 mmHg x 0,024
= 48,24 mmHg
January 29, 2008 at 9:39 PM
Nama : Vanny Adha
kelas XII IPA
absen: 40
Diket : W1 = 600 gram
W2= 27 gram
Tb = 100,13°C
Dit : Mr…………………?
Jawab :
Delta Tb = (100,13-100)°C = 0,03°C
Delta Tb = W2 / Mr x 1000 / W2 x Kb
0,13 = 27/Mrx1000/600x0,51°C/molal
0,13 mr = 27000 / 600 x 0,51°C
= 0,0176 gr/mol
January 29, 2008 at 9:40 PM
Nama : kiful minalmes
kelas XII IPA
absen: 16
Diket : W1 = 600 gram
W2= 27 gram
Tb = 100,13°C
Dit : Mr…………………?
Jawab :
Delta Tb = (100,13-100)°C = 0,03°C
Delta Tb = W2 / Mr x 1000 / W2 x Kb
0,13 = 27/Mrx1000/600x0,51°C/molal
0,13 mr = 27000 / 600 x 0,51°C
= 0,0176 gr/mol
January 29, 2008 at 9:42 PM
nama; nurfitasari
kelas; XII Ipa
Dik: Mp: 72 gram
Mrp: 18
Po: 20,1 cm Hg
Mt: 18 gram
Mrt: 180
Dit: DP
Jwb:
Mol t = gram = 18:180 = 0.1 mol
Mr
Mol p = Gram = 72 :18 = 4mol
Mr
DP = Po . Xt
= Po nt
nt + np
= 20.1 0.1
0.1 + 4
= 20.1 . 0.024
=0.482 cm Hg
pak emailku gak bs d buka. kmren dh buat tp gk bs. jd email saya gabung sm emailnya neny.maap ya pak................
January 30, 2008 at 5:38 AM
nama; nurfitasari
kelas; XII Ipa
Dik: Mp: 72 gram
Mrp: 18
Po: 20,1 cm Hg
Mt: 18 gram
Mrt: 180
Dit: DP
Jwb:
Mol t = gram = 18:180 = 0.1 mol
Mr
Mol p = Gram = 72 :18 = 4mol
Mr
DP = Po . Xt
= Po nt
nt + np
= 20.1 0.1
0.1 + 4
= 20.1 . 0.024
=0.482 cm Hg
January 30, 2008 at 5:39 AM
nama : anita andayani
kelas :i2 ipa
no.absen:04
diketahui : w1= 600 gr
w2= 27 gr
Tb= 100,13 derajat
Kb air= 0,51 derajat C/molal
dit : Mr senyawa tsb ?
jwb : delta Tb= 100,13 - 100 = 0,13
delta Tb= w2/Mr.1000/w1.Kb
0,13= (27/Mr.1000/600).0,51
0,13= 13770/600Mr
78 Mr= 13770
Mr = 13770/78
Mr = 176,6
January 30, 2008 at 5:43 AM
nama : nEnNy aPRIYantI
kelas :i2 ipa
no.absen:26
diketahui : w1= 600 gr
w2= 27 gr
Tb= 100,13 derajat
Kb air= 0,51 derajat C/molal
dit : Mr senyawa tsb ?
jwb : delta Tb= 100,13 - 100 = 0,13
delta Tb= w2/Mr.1000/w1.Kb
0,13= (27/Mr.1000/600).0,51
0,13= 13770/600Mr
78 Mr= 13770
Mr = 13770/78
Mr = 176,63
January 30, 2008 at 5:45 AM
nama; aRiZ caTUR Y
kelas; XII Ipa
no. absen: 03
Dik: Mp: 72 gram
Mrp: 18
Po: 20,1 cm Hg
Mt: 18 gram
Mrt: 180
Dit: DeLta P
Jwb:
Mol t = gram = 18:180 = 0.1 mol
Mr
Mol p = Gram = 72 :18 = 4mol
Mr
Delta P = Po . Xt
= Po nt
nt + np
= 20.1 0.1
0.1 + 4
= 20.1 . 0.024
=0.482 cm Hg
January 30, 2008 at 5:48 AM
Nama : M.Santoso
kelas XII IPA
absent:24
hal:remidi kimia
Diket : W1 = 600 gram
W2= 27 gram
Tb = 100,13°C
Dit : Mr…………………?
Jawab :
Delta Tb = (100,13-100)°C = 0,03°C
Delta Tb = W2 / Mr x 1000 / W2 x Kb
0,13 = 27/Mrx1000/600x0,51°C/molal
0,13 mr = 27000 / 600 x 0,51°C
= 0,0176 gr/mol
January 30, 2008 at 8:18 PM
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