{"id":278872,"date":"2020-01-16T11:42:04","date_gmt":"2020-01-16T04:42:04","guid":{"rendered":"https:\/\/quipperhome.wpcomstaging.com\/?p=278872"},"modified":"2020-01-16T11:44:23","modified_gmt":"2020-01-16T04:44:23","slug":"utbk-kimia-2018","status":"publish","type":"post","link":"https:\/\/quipperhome.wpcomstaging.com\/masuk-ptn\/sbmptn\/soal-sbmptn\/utbk-kimia-2018\/","title":{"rendered":"Soal UTBK SBMPTN Kimia 2018"},"content":{"rendered":"

\"\"<\/p>\n

Hai Quipperian, sudah siapkah kamu menyambut seleksi akbar, SBMPTN? Atau justru kamu masih bingung mencari soal-soal SBMPTN? Bagi kamu yang saat ini masih bingung mencari soal-soal SBMPTN, tenang Quipperian, karena kamu memilih artikel yang tepat.<\/span><\/p>\n

Kali ini, Quipper Blog akan membahas latihan soal SBMPTN Kimia TKD SAINTEK 2018. Dalam latihan soal ini, akan disertakan pula jawaban dan pembahasannya. Jadi, tenang saja ya! Yuk, <\/span>capcus <\/span><\/em>intip soal-soalnya di bawah ini.<\/span><\/p>\n

Latihan Soal 1<\/strong><\/h2>\n

Persentase massa atom oksigen\u00a0(<\/span>Ar<\/span><\/em>) =16\u00a0dalam suatu senyawa adalah\u00a024%.\u00a0Jika tetapan Avogadro\u00a0= 6,0 \u00d7 10<\/span>23<\/span><\/sup>,\u00a0jumlah atom oksigen dalam\u00a02\u00a0g\u00a0senyawa tersebut adalah…<\/span><\/p>\n

    \n
  1. 6,0 \u00d7 10<\/span>21<\/span><\/sup><\/li>\n
  2. 3,0 \u00d7 10<\/span>22<\/span><\/sup><\/li>\n
  3. 1,2 \u00d7 10<\/span>22<\/span><\/sup><\/li>\n
  4. 1,8 \u00d7 10<\/span>22<\/span><\/sup><\/li>\n
  5. 2,4 \u00d7 10<\/span>22<\/span><\/sup><\/li>\n<\/ol>\n

    Jawaban: D<\/strong><\/p>\n

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

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

      \n
    • Kadar oksigen dalam suatu senyawa\u00a0= 24%.<\/span><\/li>\n
    • Tetapan Avogadro = 6,0 \u00d7 10<\/span>23<\/span><\/sup>.<\/span><\/li>\n<\/ul>\n

      Ditanya: Jumlah atom oksigen dalam 2 gram senyawa tersebut =\u2026?<\/span><\/p>\n

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

      Misalkan jumlah atom oksigen dalam\u00a01\u00a0molekul senyawa adalah\u00a0<\/span>x<\/span><\/em>.<\/span><\/p>\n

      \"\"<\/p>\n

      Jadi, jumlah atom oksigen dalam\u00a02\u00a0gram\u00a0senyawa tersebut adalah\u00a01,8 \u00d7 10<\/span>22<\/span><\/sup>.<\/span><\/p>\n

      Latihan Soal 2<\/strong><\/h2>\n

      Nilai\u00a0pH\u00a0larutan obat batuk diatur dengan menambahkan garam yang dibuat dari asam klorida dan amonia\u00a0(<\/span>K<\/span><\/em>b<\/span><\/em><\/sub>=10<\/span>\u22125<\/span><\/sup>).\u00a0Jumlah garam yang ditambahkan ke dalam\u00a0300\u00a0mL\u00a0obat batuk tersebut agar\u00a0pH = 6\u00a0adalah…<\/span><\/p>\n

        \n
      1. 3 mmol<\/span><\/li>\n
      2. 30 mmol<\/span><\/li>\n
      3. 0,3 mmol<\/span><\/li>\n
      4. 3 mol<\/span><\/li>\n
      5. 0,3 mmol<\/span><\/li>\n<\/ol>\n

        Jawaban: E<\/strong><\/p>\n

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

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

          \n
        • pH\u00a0larutan obat batuk\u00a0= 6<\/span><\/li>\n
        • Volume obat batuk\u00a0= 300\u00a0mL<\/span><\/li>\n
        • K<\/span><\/em>b<\/span><\/em><\/sub>\u00a0amonia\u00a0(NH<\/span>3<\/span><\/sub>) = 10<\/span>\u22125<\/span><\/sup><\/li>\n<\/ul>\n

          Ditanya: Mol garam yang ditambahkan ke dalam obat batuk agar pH-nya 6 = \u2026?<\/span><\/p>\n

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

          Jika obat batuk diharapkan memiliki\u00a0pH = 6,\u00a0maka\u00a0[H<\/span>+<\/span><\/sup>]\u00a0dalam larutan tersebut adalah\u00a010<\/span>\u22126<\/span><\/sup>\u00a0M.<\/span> Agar mencapai\u00a0pH = 6\u00a0obat batuk tersebut harus ditambahkan dengan garam yang terbuat dari campuran amonia dan asam klorida (garam\u00a0NH<\/span>4<\/span><\/sub>C\u2113).\u00a0Untuk menghitung mol garam dapat menggunakan persamaan berikut.<\/span><\/p>\n

          \"\"<\/p>\n

          Jadi, mol garam yang ditambahkan ke dalam obat batuk agar memiliki\u00a0pH = 6\u00a0adalah\u00a00,3 mmol.<\/span><\/p>\n

          Latihan Soal 3<\/strong><\/h2>\n

          Unsur\u00a0F\u00a0(nomor atom\u00a0= 9) dan\u00a0M\u00a0(nomor atom\u00a0= 54) membentuk molekul\u00a0MF<\/span>4<\/span><\/sub>.\u00a0Bentuk molekul dan sifat kepolaran molekul\u00a0MF<\/span>4<\/span><\/sub>\u00a0adalah…<\/span><\/p>\n

            \n
          1. bipiramida segitiga dan polar<\/span><\/li>\n
          2. piramida dan polar<\/span><\/li>\n
          3. tetrahedral dan nonpolar<\/span><\/li>\n
          4. planar segiempat dan nonpolar<\/span><\/li>\n
          5. jungkat-jungkit dan polar<\/span><\/li>\n<\/ol>\n

            Jawaban: D<\/strong><\/p>\n

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

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

              \n
            • Nomor atom F = 9<\/span><\/li>\n
            • Nomor atom M = 54<\/span><\/li>\n<\/ul>\n

              Ditanya: Bentuk molekul dan kepolaran MF<\/span>4<\/span><\/sub> =\u2026?<\/span><\/p>\n

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

              Perhatikan konfigurasi elektron berikut.<\/span><\/p>\n

              \"\"<\/p>\n

              Kedua unsur tersebut membentuk senyawa\u00a0MF4<\/sub>.\u00a0Dalam senyawa tersebut, unsur\u00a0M\u00a0menjadi atom pusat. Hal tersebut karena\u00a0M\u00a0mengikat\u00a04\u00a0elektron dari\u00a0F,\u00a0sedangkan masing-masing unsur\u00a0F\u00a0mengikat\u00a01\u00a0elektron dari\u00a0M.\u00a0<\/span><\/p>\n

              Unsur yang mengikat elektron lebih banyak menjadi atom pusat. Bentuk molekul dari unsur tersebut dapat diketahui dari jumlah pasangan elektron ikatan\u00a0(PEI)\u00a0dan pasangan elektron bebas\u00a0(PEB).<\/span><\/p>\n

              \"\"<\/span><\/p>\n

              Oleh karena MF<\/span>4<\/span><\/sub> memiliki 4 PEI dan 2 PEB, maka senyawa tersebut memiliki bentuk molekul planar segiempat. Untuk bentuk planar segiempat, akan ditunjukkan oleh gambar berikut.<\/span><\/p>\n

              \"\"<\/p>\n

              Bentuk molekul planar segiempat merupakan bentuk yang simetris sehingga\u00a0MF<\/span>4<\/sub>\u00a0<\/span>bersifat nonpolar.<\/span><\/p>\n

              Jadi, senyawa\u00a0MF<\/span>4<\/span><\/sub>\u00a0memiliki bentuk molekul planar segiempat dan bersifat nonpolar.<\/span><\/p>\n

              Latihan Soal 4<\/strong><\/h2>\n

              Perhatikan senyawa berikut.<\/span><\/p>\n

              \"\"<\/p>\n

              Senyawa yang bersifat optis aktif ditunjukkan oleh nomor\u2026<\/span><\/p>\n

                \n
              1. 1 dan 3<\/span><\/li>\n
              2. 2, 3, dan 4<\/span><\/li>\n
              3. 2 dan 4<\/span><\/li>\n
              4. 1, 2, dan 3<\/span><\/li>\n
              5. 4<\/span><\/li>\n<\/ol>\n

                Jawaban: A<\/strong><\/p>\n

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

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

                \"\"<\/p>\n

                Ditanya: Senyawa yang bersifat optis aktif =\u2026?<\/span><\/p>\n

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

                Senyawa yang bersifat optis aktif\u00a0adalah senyawa yang dapat memutar bidang polarisasi. Suatu senyawa bersifat optis aktif jika senyawa tersebut memiliki atom\u00a0C\u00a0asimetris atau\u00a0C\u00a0kiral.<\/span> Atom C asimetris\u00a0\u00a0adalah atom\u00a0C\u00a0yang mengikat empat unsur atau molekul yang berbeda.<\/span><\/p>\n

                \"\"<\/p>\n

                Rumus struktur senyawa:<\/span><\/p>\n

                \"\"<\/p>\n

                Senyawa (1) memiliki 2 atom\u00a0C\u00a0asimetris sedangkan senyawa (2) tidak memiliki atom\u00a0C\u00a0asimetris.<\/span><\/p>\n

                \"\"<\/p>\n

                Senyawa (3) memiliki 1 atom\u00a0C\u00a0asimetris sedangkan senyawa (4) tidak memiliki atom\u00a0C\u00a0asimetris.<\/span><\/p>\n

                Jadi, senyawa yang bersifat optis aktif adalah senyawa 1 dan 3.<\/span><\/p>\n

                Latihan Soal 5<\/strong><\/h2>\n

                Energi ionisasi (kJ\/mol)\u00a0ke-1\u00a0sampai\u00a0ke-5\u00a0untuk unsur\u00a0X\u00a0berturut-turut adalah\u00a0786, 1.580, 3.230, 4.360, dan\u00a016.010.\u00a0Senyawa yang dapat terbentuk dan stabil adalah…<\/span><\/p>\n

                  \n
                1. XCl<\/span>2<\/span><\/sub><\/li>\n
                2. XCl<\/span>3<\/span><\/sub><\/li>\n
                3. XCl<\/span>4<\/span><\/sub><\/li>\n
                4. X<\/span>2<\/span><\/sub>O<\/span>3<\/span><\/sub><\/li>\n
                5. XO<\/span>3<\/span><\/sub><\/li>\n<\/ol>\n

                  Jawaban: C<\/strong><\/p>\n

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

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

                  Energi ionisasi (kJ\/mol)\u00a0ke-1\u00a0sampai\u00a0ke-5\u00a0untuk unsur\u00a0X\u00a0berturut-turut adalah\u00a0786, 1.580, 3.230, 4.360, dan\u00a016.010.<\/span><\/p>\n

                  Ditanya: \u00a0Senyawa yang dapat dibentuk dan stabil antara unsur\u00a0X\u00a0dengan Cl dan O =\u2026?<\/span><\/p>\n

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

                  Pertama, tentukan dahulu selisih energi ionisasinya.<\/span><\/p>\n

                    \n
                  • Ke-1 dan ke-2 = 1.580 \u2013 785 = 795 kJ\/mol<\/span><\/li>\n
                  • Ke-2 dan ke-3 = 3.230 \u2013 1.580 = 1.650 kJ\/mol<\/span><\/li>\n
                  • Ke-3 dan ke-4 = 4.360 \u2013 3.230 = 1.130 kJ\/mol<\/span><\/li>\n
                  • Ke-4 dan ke-5 = 16.010 \u2013 4.360 = 11.650 kJ\/mol<\/span><\/li>\n<\/ul>\n

                    Oleh karena selisih energi ionisasi paling besar ada pada energi ionisasi\u00a0ke-4\u00a0dan ke-5,\u00a0maka untuk stabil (menjadi oktet) unsur\u00a0X\u00a0akan mengikat\u00a04\u00a0elektron membentuk ion\u00a0X<\/span>4\u2212<\/span><\/sup>.<\/span><\/p>\n

                    Cara cepat untuk menentukan rumus molekul senyawa adalah sebagai berikut.<\/span><\/p>\n

                    1. Ikatan ion<\/span><\/p>\n

                    \"\"<\/p>\n

                    2. Ikatan kovalen<\/span><\/p>\n

                    \"\"<\/p>\n

                    Senyawa yang stabil dihasilkan jika berikatan dengan\u00a0Cl\u00a0(ion\u00a0Cl<\/span>–<\/span><\/sup>)\u00a0adalah sebagai berikut.<\/span><\/p>\n

                    X<\/span>4\u2212 <\/span><\/sup>+ Cl<\/span>\u2212 <\/span><\/sup>\u2192 XCl<\/span>4<\/span><\/sub><\/p>\n

                    Senyawa stabil yang dihasilkan jika berikatan dengan\u00a0O (ion O<\/span>2\u2212<\/span>)\u00a0adalah sebagai berikut.<\/span><\/p>\n

                    X<\/span>4\u2212<\/sup> <\/span>+ O<\/span>2<\/sup>\u2212<\/span> \u2192 X<\/span>2<\/span><\/sub>O<\/span>4<\/span><\/sub> \u2192 XO<\/span>2<\/span><\/sub><\/p>\n

                    Jadi, senyawa yang dapat dibentuk dan stabil antara unsur\u00a0X\u00a0dengan\u00a0Cl\u00a0dan\u00a0O\u00a0adalah\u00a0XCl<\/span>4<\/span><\/sub>\u00a0dan\u00a0XO<\/span>2<\/span><\/sub>.<\/span><\/p>\n

                    Bagaimana Quipperian, apakah kamu sudah mulai paham mengerjakan latihan soal SBMPTN Kimia TKD SAINTEK 2018 di atas? Agar pemahamanmu semakin terasah, sering-seringlah mengerjakan latihan soal. Ingat, selangkah lagi kamu akan memasuki PTN impian.\u00a0<\/span><\/p>\n

                    Untuk mendukung kesuksesanmu masuk PTN, Quipper Video hadir dengan Paket Intensif UTBK SBMPTN 2020. Semua kebutuhan sudah tersedia di dalamnya, mulai dari materi persiapan, materi pemantapan, latihan soal, bank soal, hingga pembahasan lengkapnya. Jadi, tunggu apalagi. Yuk, gabung Quipper Video sekarang!<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

                    Hai Quipperian, sudah siapkah kamu menyambut seleksi akbar, SBMPTN? Atau justru kamu masih bingung mencari soal-soal SBMPTN? Bagi kamu yang saat ini masih bingung…<\/p>\n","protected":false},"author":99400369,"featured_media":278890,"comment_status":"closed","ping_status":"closed","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_feature_clip_id":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[679385273,679385209],"tags":[679385465],"ppma_author":[679386827],"class_list":["post-278872","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-sbmptn","category-soal-sbmptn","tag-soal-utbk-saintek-2018"],"acf":[],"yoast_head":"\nSoal UTBK SBMPTN Kimia 2018 - Quipper Blog<\/title>\n<meta name=\"description\" content=\"Galau karena enggak menemukan kumpulan soal Kimia SBMPTN 2018? Tenang, ada di sini, kok! 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