Disoluciones.ipynb

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        "<a href=\"https://colab.research.google.com/github/Polux79/Ingenieria-ambiental/blob/main/Disoluciones.ipynb\" target=\"_parent\"><img src=\"https://colab.research.google.com/assets/colab-badge.svg\" alt=\"Open In Colab\"/></a>"
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        "\n",
        "\n",
        "1.  Calcule la masa de $KI$, en gramos, que se requiere para preparar $5,00 x 10^2$ $mL$ de una disolución $2,80M$."
      ],
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          "text": [
            "232.40 gramos de KI\n"
          ]
        }
      ],
      "source": [
        "#Primera forma de solucionar, ingreso detallado de variables ---> Nivel principiante\n",
        "\n",
        "#Calculo del peso molecular del KI\n",
        "Peso_K = 39.0983      #De la tabla periódica, Peso atómico (P.A)\n",
        "Peso_I = 126.9045     #De la tabla periódica, P.A\n",
        "\n",
        "Peso_KI = Peso_K + Peso_I   # El peso molecular del KI, suma de P.A\n",
        "\n",
        "# print (Peso_KI) #si deseas visualizar el valor de peso molecular elimina el # del inicio de linea\n",
        "\n",
        "#Transformando el volumen a Litros\n",
        "V_sol = 5*10**2*(1/1000) #mL\n",
        "\n",
        "#Declaración de la molaridad\n",
        "M_sol = 2.8     #M\n",
        "\n",
        "#Calculo de la masa de KI\n",
        "m_KI = Peso_KI*V_sol*M_sol\n",
        "print(f\"{m_KI:.2f} gramos de KI\")"
      ]
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      "cell_type": "code",
      "source": [
        "#Segunda forma de solucionar, uso de la función \"def\" ---> Nivel avanzado\n",
        "\n",
        "def masa_soluto(Molaridad, Peso_molecular, Volumen):\n",
        "\n",
        "    return Molaridad * Peso_molecular * Volumen\n",
        "\n",
        "resultado = masa_soluto(2.8, (39.0983+126.9045), (500/1000)) # Las variables se transforman al interior de los ()\n",
        "\n",
        "print(f\"{resultado:.2f} gramos de KI\")"
      ],
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      "outputs": [
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          "text": [
            "232.40 gramos de KI\n"
          ]
        }
      ]
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