{"id":279237,"date":"2020-01-23T13:38:44","date_gmt":"2020-01-23T06:38:44","guid":{"rendered":"https:\/\/quipperhome.wpcomstaging.com\/?p=279237"},"modified":"2020-01-23T17:34:30","modified_gmt":"2020-01-23T10:34:30","slug":"utbk-kimia-2017","status":"publish","type":"post","link":"https:\/\/quipperhome.wpcomstaging.com\/masuk-ptn\/sbmptn\/soal-sbmptn\/utbk-kimia-2017\/","title":{"rendered":"Soal UTBK SBMPTN Kimia 2017"},"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, tampaknya kamu berada di artikel yang tepat, <\/span>nih<\/span><\/em>.\u00a0<\/span><\/p>\n

Kali ini, Quipper Blog akan membahas latihan soal SBMPTN Kimia TKD SAINTEK 2017. Kabar baiknya, latihan soal yang disediakan Quipper Blog, lengkap dengan jawaban beserta pembahasannya, <\/span>lho<\/span><\/em>. Semakin penasaran? Yuk, segera kerjakan soalnya!<\/span><\/p>\n

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

Senyawa kovalen\u00a0X<\/span>2<\/span><\/sub>Y\u00a0terbentuk dari atom dengan nomor atom X dan Y berturut-turut adalah\u00a017\u00a0dan\u00a08.\u00a0Bentuk molekul yang sesuai untuk senyawa kovalen tersebut adalah…<\/span><\/p>\n

    \n
  1. piramida segitiga<\/span><\/li>\n
  2. linear<\/span><\/li>\n
  3. tetrahedral<\/span><\/li>\n
  4. bentuk V<\/span><\/li>\n
  5. segitiga datar<\/span><\/li>\n<\/ol>\n

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

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

    Bentuk molekul dapat ditentukan dengan menggunakan teori domain elektron. Mula-mula, tentukan dahulu konfigurasi atom X dan Y.<\/span><\/p>\n

    \"\"<\/p>\n

    Saat berikatan menjadi\u00a0X<\/span>2<\/span>Y,\u00a0atom pusatnya adalah Y.<\/span><\/p>\n

    \"\"<\/p>\n

    Senyawa ini memiliki domain elektron\u00a08\u00a0(4\u00a0pasang), yaitu\u00a02\u00a0pasang merupakan pasangan elektron ikatan (PEI) dan\u00a02\u00a0pasang merupakan pasangan elektron bebas (PEB). Dengan demikian, bentuk molekulnya adalah bentuk V.<\/span><\/p>\n

    Jadi, bentuk molekul yang sesuai untuk senyawa kovalen\u00a0X<\/span>2<\/span><\/sub>Y\u00a0adalah bentuk V.<\/span><\/p>\n

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

    Diberikan persamaan reaksi sebagai berikut.<\/span><\/p>\n

    \"\"<\/p>\n

    Suatu bahan yang mengandung\u00a060%\u00a0massa\u00a0NaBH<\/span>4<\/span><\/sub>\u00a0(Mr=38)\u00a0digunakan dalam reaksi ini. Massa bahan tersebut yang diperlukan untuk menghasilkan\u00a00,20\u00a0mol\u00a0B<\/span>2<\/span><\/sub>H<\/span>6<\/sub>\u00a0<\/span>adalah…<\/span><\/p>\n

      \n
    1. 6,8 gram<\/span><\/li>\n
    2. 19 gram<\/span><\/li>\n
    3. 22,8 gram<\/span><\/li>\n
    4. 38,0 gram<\/span><\/li>\n
    5. 11,4 gram<\/span><\/li>\n<\/ol>\n

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

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

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

      Mol B<\/span>2<\/span><\/sub>H<\/span>6<\/span><\/sub> = 0,2<\/span><\/p>\n

      Mr<\/span><\/em> NaBH<\/span>4<\/span><\/sub> = 38<\/span><\/p>\n

      Ditanya: Massa bahan =\u2026?<\/span><\/p>\n

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

      Reaksi pada soal sudah setara, sehingga Quipperian tidak perlu menyetarakannya kembali.<\/span><\/p>\n

      \"\"<\/p>\n

      Oleh karena pada reaksi setara perbandingan koefisien sama dengan perbandingan mol, maka:<\/span><\/p>\n

      \"\"<\/p>\n

      Ini berarti, massa NaBH<\/span>4<\/span> dalam reaksi tersebut adalah sebagai berikut.<\/span><\/p>\n

      \"\"<\/p>\n

      Dengan demikian, massa bahan yang digunakan adalah sebagai berikut.<\/span><\/p>\n

      \"\"<\/p>\n

      Jadi, massa bahan yang diperlukan adalah 19 gram.<\/span><\/p>\n

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

      Dalam sebuah percobaan elektrolisis, digunakan dua sel yang dirangkai secara seri. Masing-masing sel menerima arus listrik yang sama. Sel pertama berisi AgNO<\/span>3<\/span><\/sub>, sedangkan sel kedua berisi XCl<\/span>3<\/span><\/sub>. Jika setelah elektrolisis selesai diperoleh\u00a01,44\u00a0gram logam Ag pada sel pertama dan\u00a00,12\u00a0logam X pada sel kedua, massa molar logam X adalah … (<\/span>Ar<\/span><\/em> Ag =\u00a0108)<\/span><\/p>\n

        \n
      1. 54<\/span><\/li>\n
      2. 30<\/span><\/li>\n
      3. 36<\/span><\/li>\n
      4. 32<\/span><\/li>\n
      5. 28<\/span><\/li>\n<\/ol>\n

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

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

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

        Massa Ag = 1,44 g<\/span><\/p>\n

        Massa X = 0,12 g<\/span><\/p>\n

        Ditanya: <\/span>Ar<\/span><\/em> X =\u2026?<\/span><\/p>\n

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

        Sel pertama:<\/span><\/p>\n

        \"\"<\/p>\n

        Ini berarti:<\/span><\/p>\n

        \"\"<\/p>\n

        Sel kedua (XCl<\/span>3<\/span><\/sub>):<\/span><\/p>\n

        \"\"<\/p>\n

        Oleh karena digunakan arus yang sama, maka mol elektron yang digunakan juga sama, yaitu 0,013.<\/span><\/p>\n

        \"\"<\/p>\n

        Dengan demikian, diperoleh:<\/span><\/p>\n

        \"\"<\/p>\n

        Jadi, massa molar logam X adalah\u00a030.<\/span><\/p>\n

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

        Suatu senyawa organik berwujud cair pada suhu kamar, berbau khas, dengan pereaksi Tollens tidak menghasilkan endapan mengkilat menyerupai cermin, jika direduksi menghasilkan suatu alkohol sekunder, dan sering digunakan sebagai pelarut dalam industri. Senyawa organik tersebut adalah…<\/span><\/p>\n

          \n
        1. etanol<\/span><\/li>\n
        2. n<\/span><\/em>-heksana<\/span><\/li>\n
        3. etil etanoat<\/span><\/li>\n
        4. propanon<\/span><\/li>\n
        5. benzaldehida<\/span><\/li>\n<\/ol>\n

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

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

          Kata kunci dari soal ini adalah sebagai berikut.<\/span><\/p>\n

            \n
          • Tidak menghasilkan endapan dengan pereaksi Tollens.<\/span><\/li>\n
          • Jika direduksi menghasilkan alkohol sekunder.<\/span><\/li>\n
          • Digunakan sebagai pelarut dalam industri.<\/span><\/li>\n<\/ul>\n

            Dari kata kunci kedua, dapat diketahui bahwa senyawa yang direduksi menghasilkan alkohol sekunder adalah senyawa dari golongan keton atau alkanon yang mengandung gugus (-CO-).<\/span><\/p>\n

            Jadi, senyawa organik tersebut adalah propanon.<\/span><\/p>\n

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

            Gas oksigen difluorida\u00a0(OF<\/span>2<\/span><\/sub>)\u00a0disintesis dari reaksi antara gas\u00a0F<\/span>2<\/sub>\u00a0<\/span>dan\u00a0O<\/span>2<\/span><\/sub>\u00a0menurut reaksi berikut.<\/span><\/p>\n

            \"\"<\/p>\n

            Dalam sebuah wadah dengan volume tertentu, tekanan awal gas\u00a0F<\/span>2<\/span><\/sub>\u00a0dan\u00a0O<\/span>2<\/span><\/sub>\u00a0diketahui masing-masing adalah\u00a01\u00a0atm. Jika pada kesetimbangan tekanan total gas adalah\u00a01,75\u00a0atm, nilai\u00a0<\/span>K<\/span><\/em>p<\/span><\/sub>\u00a0reaksi tersebut adalah…<\/span><\/p>\n

              \n
            1. 1,333<\/span><\/li>\n
            2. 0,278<\/span><\/li>\n
            3. 0,755<\/span><\/li>\n
            4. 0,133<\/span><\/li>\n
            5. 0,555<\/span><\/li>\n<\/ol>\n

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

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

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

              Gas oksigen difluorida\u00a0(OF<\/span>2<\/span><\/sub>)\u00a0disintesis dari reaksi antara gas\u00a0F<\/span>2\u00a0<\/span><\/sub>dan\u00a0O<\/span>2<\/span><\/sub>\u00a0menurut reaksi berikut.<\/span><\/p>\n

              \"\"<\/p>\n

              Dalam sebuah wadah dengan volume tertentu, tekanan awal gas\u00a0F<\/span>2<\/span><\/sub>\u00a0dan\u00a0O<\/span>2<\/span><\/sub>\u00a0diketahui masing-masing adalah\u00a01\u00a0atm. Pada kesetimbangan tekanan total gas adalah\u00a01,75\u00a0atm.<\/span><\/p>\n

              Ditanya: Nilai <\/span>K<\/span><\/em>p<\/span><\/sub> =\u2026?<\/span><\/p>\n

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

              Reaksi kesetimbangan yang terjadi adalah sebagai berikut.<\/span><\/p>\n

              \"\"<\/p>\n

              Oleh karena tekanan total saat setimbang adalah 1,75 atm, maka:<\/span><\/p>\n

              \"\"<\/p>\n

              Ini berarti, tekanan masing-masing zat saat setimbang adalah sebagai berikut.<\/span><\/p>\n

              \"\"<\/p>\n

              Dengan demikian, nilai <\/span>K<\/span><\/em>p<\/span><\/em><\/sub> reaksi tersebut adalah sebagai berikut.<\/span><\/p>\n

              \"\"<\/p>\n

              Jadi, nilai <\/span>K<\/span><\/em>p<\/span><\/em><\/sub> reaksi tersebut adalah 1,333.<\/span><\/p>\n

              Latihan Soal 6<\/strong><\/h2>\n

              Suatu larutan buffer dibuat dengan cara mencampurkan\u00a00,6\u00a0mol asam asetat dan\u00a00,2\u00a0mol NaOH dalam\u00a0500\u00a0mL larutan\u00a0(<\/span>K<\/span><\/em>a<\/span><\/em><\/sub> CH<\/span>3<\/span><\/sub>COOH = 5 \u00d7 10<\/span>\u22125<\/span><\/sup>).\u00a0pH larutan tersebut adalah…<\/span><\/p>\n

                \n
              1. 6<\/span><\/li>\n
              2. 4<\/span><\/li>\n
              3. 3<\/span><\/li>\n
              4. 5<\/span><\/li>\n
              5. 2<\/span><\/li>\n<\/ol>\n

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

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

                Diketahui:<\/span>
                \n<\/span>mol CH<\/span>3<\/span><\/sub>COOH =\u00a00,6\u00a0mol<\/span>
                \n<\/span>mol NaOH =\u00a00,2\u00a0mol<\/span>
                \n<\/span>volume larutan =\u00a0500\u00a0mL<\/span><\/p>\n

                Ditanya: pH = … ?<\/span><\/p>\n

                Penyelesaian:<\/span>
                \n<\/span>Reaksi yang terjadi adalah sebagai berikut.<\/span><\/p>\n

                \"\"<\/p>\n

                Oleh karena yang tersisa adalah asam lemah, maka larutan tersebut merupakan larutan penyangga asam yang pH-nya dapat dihitung dengan cara berikut.<\/span><\/p>\n

                \"\"<\/p>\n

                Dengan demikian, diperoleh:<\/span><\/p>\n

                \"\"<\/p>\n

                Jadi, pH larutan tersebut adalah\u00a04.<\/span><\/p>\n

                Latihan Soal 7<\/strong><\/h2>\n

                Setelah disimpan selama\u00a0120\u00a0hari, suatu unsur radioaktif masih tersisa sebanyak\u00a012,5%\u00a0dari jumlah semula. Waktu paruh unsur tersebut adalah…<\/span><\/p>\n

                  \n
                1. 60 hari<\/span><\/li>\n
                2. 20 hari<\/span><\/li>\n
                3. 40 hari<\/span><\/li>\n
                4. 30 hari<\/span><\/li>\n
                5. 9,6 hari<\/span><\/li>\n<\/ol>\n

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

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

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

                  N<\/span><\/em>t<\/span><\/em><\/sub> = 12,5% = 0,125<\/span>N<\/span><\/em>0<\/span><\/em><\/sub><\/p>\n

                  t<\/span><\/em> = 120 hari<\/span><\/p>\n

                  Ditanya: <\/span>t<\/span><\/em>1\/2<\/span><\/em><\/sub> =\u2026?<\/span><\/p>\n

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

                  Waktu paruh dapat ditentukan dengan rumus berikut.<\/span><\/p>\n

                  \"\"<\/p>\n

                  Jadi, waktu paruh unsur tersebut adalah 40 hari.<\/span><\/p>\n

                  Bagaimana Quipperian, apakah kamu sudah mulai paham mengerjakan latihan soal SBMPTN Kimia TKD SAINTEK 2017 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":279264,"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":[679385508],"ppma_author":[679386827],"class_list":["post-279237","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-sbmptn","category-soal-sbmptn","tag-soal-utbk-saintek-2017"],"acf":[],"yoast_head":"\nSoal UTBK SBMPTN Kimia 2017 - Quipper Blog<\/title>\n<meta name=\"description\" content=\"Bagi kamu yang sedang mencari-cari soal Kimia SBMPTN 2017, yuk mampir ke sini! 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