{"id":16112,"date":"2016-01-22T18:06:10","date_gmt":"2016-01-22T11:06:10","guid":{"rendered":"https:\/\/video.quipper.com\/id\/blog\/?p=16112"},"modified":"2021-11-24T22:29:54","modified_gmt":"2021-11-24T15:29:54","slug":"yuk-belajar-pembahasan-soal-sbmptn-kimia-agar-lulus-dengan-nilai-tinggi","status":"publish","type":"post","link":"https:\/\/quipperhome.wpcomstaging.com\/masuk-ptn\/sbmptn\/kimia-sbmptn\/yuk-belajar-pembahasan-soal-sbmptn-kimia-agar-lulus-dengan-nilai-tinggi\/","title":{"rendered":"Yuk Belajar Pembahasan Soal SBMPTN Kimia Agar Lulus dengan Nilai Tinggi!"},"content":{"rendered":"

\"<\/a><\/p>\n

Quipperian, untuk kamu yang memilih program studi di bidang Saintek, pastikan kamu mempersiapkan diri dengan baik untuk menghadapi Tes Kemampuan Dasar (TKD) Saintek ya. Salah satu mata uji dalam TKD Saintek adalah Kimia, Quipperian. Supaya kamu semakin bersiap diri, yuk simak contoh soal dan pembahasan mata uji Kimia dalam SBMPTN berikut:<\/span><\/p>\n

 1) Gas etana C<\/i>2<\/i>H<\/i>6<\/i> (Mr = 30) sebanyak 60 gram direaksikan dengan gas bromin (Br<\/i>2<\/i>) hingga gas etana habis dan ternyata hanya menghasilkan 282 gram C<\/i>2<\/i>H<\/i>4<\/i>Br<\/i>2<\/i> (Mr = 188) dan x gram C<\/i>2<\/i>H<\/i>2<\/i>Br<\/i>4<\/i> (Mr = 346). Massa C<\/i>2<\/i>H<\/i>2<\/i>Br<\/i>4<\/i> yang dihasilkan adalah\u2026<\/i><\/strong><\/h3>\n

(A)      34,6 gram<\/span><\/p>\n

(B)     <\/span> 79,2 gram<\/span><\/p>\n

(C)      173 gram<\/span><\/p>\n

(D)      188 gram<\/span><\/p>\n

(E)     <\/span> 346 gram<\/span><\/p>\n

Jawaban:<\/span> C<\/b><\/p>\n

Pembahasan :<\/span><\/p>\n

Reaksi 1     <\/span> :  C<\/span>2<\/span>H<\/span>6<\/span> + Br<\/span>2<\/span> \u2192 C<\/span>2<\/span>H<\/span>4<\/span>Br<\/span>2<\/span> + H<\/span>2<\/span><\/p>\n

Reaksi 1     <\/span> :  C<\/span>2<\/span>H<\/span>6<\/span> + 2 Br<\/span>2<\/span> \u2192 C<\/span>2<\/span>H<\/span>2<\/span>Br<\/span>4<\/span> + 2H<\/span>2<\/span><\/p>\n

mol C<\/span>2<\/span>H<\/span>4<\/span>Br<\/span>2<\/span> = 282\/188 = 1,5 mol \u2192 mol C<\/span>2<\/span>H<\/span>6<\/span> = 1,5 mol<\/span><\/p>\n

mol C<\/span>2<\/span>H<\/span>2<\/span>Br<\/span>4<\/span> = x\/346 mol \u2192 mol C<\/span>2<\/span>H<\/span>6<\/span> = x\/346 mol<\/span><\/p>\n

mol C<\/span>2<\/span>H<\/span>6<\/span> total = 60\/30 = 2 mol<\/span><\/p>\n

2 = 1,5 + (x\/346)<\/span><\/p>\n

0,5 = x\/346<\/span><\/p>\n

x = 173 gram<\/span><\/p>\n

Maka massa C<\/span>2<\/span>H<\/span>2<\/span>Br<\/span>4<\/span> yang dihasilkan adalah <\/span>173 gram!<\/span><\/p>\n

 2. Gula 0,1 mol dan 0,1 mol garam LX dengan derajat ionisasi 0,5 masing masing dilarutkan dalam 1 liter air (\u03c1 = 1 g\/mL). Jika penurunan titik beku larutan gula t <\/i>o<\/i>C, maka penurunan titik beku larutan garam LX adalah\u2026<\/i><\/strong><\/h3>\n

(A)      0,25 t <\/span>o<\/span>C<\/span><\/p>\n

(B)     <\/span> 0,5 t <\/span>o<\/span>C<\/span><\/p>\n

(C)     <\/span> 0,75 t <\/span>o<\/span>C<\/span><\/p>\n

(D)      t <\/span>o<\/span>C<\/span><\/p>\n

(E)     <\/span> 1,5 t <\/span>o<\/span>C<\/span><\/p>\n

Jawaban: <\/span>E<\/b><\/p>\n

Pembahasan :<\/span><\/p>\n

Gula<\/span><\/p>\n

\u0394T<\/span>f<\/span>  = m.K<\/span>f<\/span><\/p>\n

t = (0,1\/1). K<\/span>f<\/span><\/p>\n

K<\/span>f<\/span> = t\/0,1<\/span><\/p>\n

Garam<\/span><\/p>\n

\u0394T<\/span>f<\/span>  = m.K<\/span>f<\/span>.i<\/span><\/p>\n

= (0,1\/1)(t\/0,1)[1+(2-1)0,5]<\/span><\/p>\n

= <\/span>1,5 t <\/span>o<\/span>C<\/span><\/p>\n

Mau Lepas SNMPTN? Kesempatan Tidak Datang Dua Kali Lho!<\/a><\/h3>\n

 3. Potensial reduksi standar E<\/i>o<\/i><\/strong><\/h3>\n

Hg<\/span><\/i>2+<\/span><\/i>(aq) + 2 e<\/span><\/i>–<\/span><\/i> \u2192 Hg(l) E<\/span><\/i>o<\/span><\/i> = 0,86 V<\/span><\/i><\/p>\n

Co<\/span><\/i>3+<\/span><\/i>(aq) + e<\/span><\/i>–<\/span><\/i> \u2192 Co<\/span><\/i>2+<\/span><\/i>(aq) E<\/span><\/i>o<\/span><\/i> = 1,82 V<\/span><\/i><\/p>\n

Harga E<\/span><\/i>o<\/span><\/i> reaksi<\/span><\/i><\/p>\n

Hg(l) + Co<\/span><\/i>3+<\/span><\/i>(aq) \u2192 Hg<\/span><\/i>2+<\/span><\/i>(aq) + Co<\/span><\/i>2+<\/span><\/i>(aq) adalah\u2026<\/span><\/i><\/p>\n

(A)        0,96 V<\/span><\/p>\n

(B)     <\/span> 1,77 V<\/span><\/p>\n

(C)     <\/span> 2,68 V<\/span><\/p>\n

(D)        2,78 V<\/span><\/p>\n

(E)     <\/span> 3,6 V<\/span><\/p>\n

Jawaban: <\/span>A<\/b><\/p>\n

Pembahasan :<\/span><\/p>\n

E<\/span>o<\/span> reaksi = E<\/span>o<\/span> red – E<\/span>o<\/span> oks<\/span><\/p>\n

= 1,82 \u2013 0,86 = <\/span>0,96 V<\/span><\/p>\n

 4. X dan Y adalah senyawa karbon. X bereaksi dengan natrium membentuk gas hydrogen. X dan Y dapat bereaksi membentuk ester. X dapat juga menjadi Y jika direaksikan dengan K<\/i>2<\/i>Cr<\/i>2<\/i>O<\/i>7<\/i> dalam media asam. Senyawa X adalah\u2026<\/i><\/h3>\n

(A)        CH<\/span>3<\/span>CH<\/span>2<\/span>COOH<\/span><\/p>\n

(B)     <\/span> CH<\/span>3<\/span>CH<\/span>2<\/span>OCH<\/span>3<\/span><\/p>\n

(C)        CH<\/span>3<\/span>CH<\/span>2<\/span>CH<\/span>2<\/span>OH<\/span><\/p>\n

(D)        CH<\/span>3<\/span>CH<\/span>2<\/span>CH<\/span>2<\/span>O<\/span><\/p>\n

(E)     <\/span> CH<\/span>3<\/span>CH<\/span>2<\/span>COCH<\/span>3<\/span>  <\/span><\/p>\n

Jawaban: <\/span>C<\/b><\/p>\n

Pembahasan :<\/span><\/p>\n

X dan Y dapat bereaksi membentuk ester \u2192 reaksi esterifikasi \u2192 Alkohol dengan Asam karboksilat membentuk ester<\/span><\/p>\n

X dapat juga menjadi Y jika direaksikan dengan K<\/span>2<\/span>Cr<\/span>2<\/span>O<\/span>7<\/span> dalam media asam. \u2192 K<\/span>2<\/span>Cr<\/span>2<\/span>O<\/span>7<\/span> adalah oksidator kuat, jika suatu alkohol primer dioksidasi maka akan menghasilkan asam karboksilat.<\/span><\/p>\n

Sehingga senyawa X adalah <\/span>suatu alkohol primer (CH<\/span>3<\/span>CH<\/span>2<\/span>CH<\/span>2<\/span>OH)<\/span><\/p>\n

\n

 <\/p>\n

Begitulah contoh soal dan pembahasan mata uji Kimia dalam SBMPTN, Quipperian. Semoga kamu terbantu dengan informasi di atas, sehingga kamu semakin siap dalam menghadapi SBMPTN ya!<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

Quipperian, untuk kamu yang memilih program studi di bidang Saintek, pastikan kamu mempersiapkan diri dengan baik untuk menghadapi Tes Kemampuan Dasar (TKD) Saintek ya.…<\/p>\n","protected":false},"author":156446345,"featured_media":16122,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"_crdt_document":"","_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[549090506],"tags":[],"ppma_author":[679386822],"class_list":["post-16112","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-kimia-sbmptn"],"acf":[],"yoast_head":"\nPembahasan Soal SBMPTN Kimia<\/title>\n<meta name=\"description\" content=\"Salah satu mata uji dalam TKD Saintek adalah Kimia, Quipperian. 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