{"id":16590,"date":"2025-06-11T16:36:26","date_gmt":"2025-06-11T16:36:26","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=16590"},"modified":"2025-06-11T16:36:32","modified_gmt":"2025-06-11T16:36:32","slug":"calculate-the-mass-in-kg-of-1-90-l-of-a-10-mol-l-agclo4","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/calculate-the-mass-in-kg-of-1-90-l-of-a-10-mol-l-agclo4\/","title":{"rendered":"Calculate the mass in kg of 1.90 L of a 10 mol\/L AgClO4"},"content":{"rendered":"\n

Calculate the mass in kg of 1.90 L of a 10 mol\/L AgClO4<\/p>\n\n\n\n

The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n

To calculate the mass in kilograms (kg)<\/strong> of 1.90 L<\/strong> of a 10 mol\/L solution of AgClO\u2084 (silver perchlorate)<\/strong>, we follow these steps:<\/p>\n\n\n\n

\n\n\n\n

Step 1: Determine the number of moles<\/strong><\/h3>\n\n\n\n

We are given:<\/p>\n\n\n\n

    \n
  • Volume = 1.90 L<\/li>\n\n\n\n
  • Molarity (M) = 10 mol\/L<\/li>\n<\/ul>\n\n\n\n

    Moles=Molarity\u00d7Volume=10\u2009mol\/L\u00d71.90\u2009L=19.0\u2009mol\\text{Moles} = Molarity \\times Volume = 10 \\, \\text{mol\/L} \\times 1.90 \\, \\text{L} = 19.0 \\, \\text{mol}<\/p>\n\n\n\n

    \n\n\n\n

    Step 2: Calculate the molar mass of AgClO\u2084<\/strong><\/h3>\n\n\n\n

    Breakdown:<\/p>\n\n\n\n

      \n
    • Ag = 107.87 g\/mol<\/li>\n\n\n\n
    • Cl = 35.45 g\/mol<\/li>\n\n\n\n
    • O\u2084 = 4 \u00d7 16.00 = 64.00 g\/mol<\/li>\n<\/ul>\n\n\n\n

      Molar mass of AgClO\u2084=107.87+35.45+64.00=207.32\u2009g\/mol\\text{Molar mass of AgClO\u2084} = 107.87 + 35.45 + 64.00 = 207.32 \\, \\text{g\/mol}<\/p>\n\n\n\n

      \n\n\n\n

      Step 3: Calculate the mass in grams<\/strong><\/h3>\n\n\n\n

      Mass=Moles\u00d7Molar mass=19.0\u2009mol\u00d7207.32\u2009g\/mol=3939.08\u2009g\\text{Mass} = \\text{Moles} \\times \\text{Molar mass} = 19.0 \\, \\text{mol} \\times 207.32 \\, \\text{g\/mol} = 3939.08 \\, \\text{g}<\/p>\n\n\n\n

      \n\n\n\n

      Step 4: Convert grams to kilograms<\/strong><\/h3>\n\n\n\n

      Mass in kg=3939.08\u2009g1000=3.93908\u2009kg\\text{Mass in kg} = \\frac{3939.08 \\, \\text{g}}{1000} = 3.93908 \\, \\text{kg}<\/p>\n\n\n\n

      \n\n\n\n

      \u2705 Final Answer: 3.94 kg (rounded to 3 significant figures)<\/strong><\/h3>\n\n\n\n
      \n\n\n\n

      300-Word Explanation<\/strong><\/h3>\n\n\n\n

      To calculate the mass of a chemical in a solution, we need to understand the concept of molarity and the molar mass of the compound. Molarity (mol\/L) expresses how many moles of solute (in this case, silver perchlorate or AgClO\u2084) are present in one liter of solution. The problem gives a molarity of 10 mol\/L and a volume of 1.90 L, meaning we can calculate the total number of moles of AgClO\u2084 using the formula: moles=molarity\u00d7volume\\text{moles} = \\text{molarity} \\times \\text{volume}<\/p>\n\n\n\n

      This gives 19.0 moles of AgClO\u2084 in 1.90 L of solution.<\/p>\n\n\n\n

      Next, we need to determine how much this quantity of AgClO\u2084 weighs. For this, we need its molar mass<\/strong>\u2014which is the mass of one mole of the compound. We calculate this by summing the atomic masses of all the atoms in the molecule:<\/p>\n\n\n\n

        \n
      • Silver (Ag) contributes 107.87 g\/mol<\/li>\n\n\n\n
      • Chlorine (Cl) adds 35.45 g\/mol<\/li>\n\n\n\n
      • Four oxygens (O\u2084) contribute 64.00 g\/mol
        Adding these gives a total molar mass of 207.32 g\/mol.<\/li>\n<\/ul>\n\n\n\n

        We multiply the molar mass by the number of moles to get the mass in grams: 19.0\u00d7207.32=3939.08 g19.0 \\times 207.32 = 3939.08 \\text{ g}<\/p>\n\n\n\n

        Finally, to convert grams to kilograms, we divide by 1000. This gives approximately 3.94 kg<\/strong>, which is the total mass of AgClO\u2084 in 1.90 L of this solution.<\/p>\n","protected":false},"excerpt":{"rendered":"

        Calculate the mass in kg of 1.90 L of a 10 mol\/L AgClO4 The correct answer and explanation is: To calculate the mass in kilograms (kg) of 1.90 L of a 10 mol\/L solution of AgClO\u2084 (silver perchlorate), we follow these steps: Step 1: Determine the number of moles We are given: Moles=Molarity\u00d7Volume=10\u2009mol\/L\u00d71.90\u2009L=19.0\u2009mol\\text{Moles} = Molarity […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-16590","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16590","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=16590"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16590\/revisions"}],"predecessor-version":[{"id":16592,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16590\/revisions\/16592"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=16590"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=16590"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=16590"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}