{"id":222981,"date":"2025-06-01T06:48:05","date_gmt":"2025-06-01T06:48:05","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=222981"},"modified":"2025-06-01T06:48:07","modified_gmt":"2025-06-01T06:48:07","slug":"the-technician-began-his-analysis-by-preparing-a-stock-solution-of-fluorescein","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/01\/the-technician-began-his-analysis-by-preparing-a-stock-solution-of-fluorescein\/","title":{"rendered":"The technician began his analysis by preparing a stock solution of fluorescein."},"content":{"rendered":"\n

The technician began his analysis by preparing a stock solution of fluorescein. To prepare the fluorescein stock solution, 0.01140 g of disodium fluorescein salt (MM = 376.28 g\/mol) was weighed on the microanalytical balance. The salt was dissolved in 50 mL of 0.1 M NaOH and then analytically transferred to a 100.00 mL volumetric flask. The flask was diluted to the mark with 0.1 M NaOH. Calculate the concentration of fluorescein (MM = 332.21 g\/mol) stock solution in ppm.
The technician began his analysis by preparing a stock solution of fluorescein. To prepare the fluorescein stock solution, 0.01140 g of disodium fluorescein salt (MM = 376.28 g\/mol) was weighed on the microanalytical balance. The salt was dissolved in 50 mL of 0.1 M NaOH and then analytically transferred to a 100.00 mL volumetric flask. The flask was diluted to the mark with 0.1 M NaOH. Calculate the concentration of fluorescein (MM = 332.21 g\/mol) stock solution in ppm.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Answer:<\/strong><\/h3>\n\n\n\n

The concentration of fluorescein<\/strong> in the final stock solution<\/strong> is 100.6 ppm<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

Step-by-Step Calculation and Explanation (\u2248300 words):<\/strong><\/h3>\n\n\n\n

To calculate the concentration in ppm (parts per million)<\/strong> of fluorescein in the stock solution, we follow these steps:<\/p>\n\n\n\n

\n\n\n\n

1. Determine moles of disodium fluorescein salt used:<\/strong><\/h3>\n\n\n\n

Given:<\/p>\n\n\n\n

    \n
  • Mass of disodium fluorescein salt = 0.01140 g<\/li>\n\n\n\n
  • Molar mass (MM) of disodium fluorescein salt = 376.28 g\/mol<\/li>\n<\/ul>\n\n\n\n

    Moles of disodium salt=0.01140 g376.28 g\/mol=3.03\u00d710\u22125 mol\\text{Moles of disodium salt} = \\frac{0.01140 \\text{ g}}{376.28 \\text{ g\/mol}} = 3.03 \\times 10^{-5} \\text{ mol}<\/p>\n\n\n\n

    \n\n\n\n

    2. Convert to moles of fluorescein:<\/strong><\/h3>\n\n\n\n

    Disodium fluorescein salt dissociates to form fluorescein (C\u2082\u2080H\u2081\u2080O\u2085), so the mole ratio is 1:1<\/strong>.
    Thus, moles of fluorescein = 3.03 \u00d7 10\u207b\u2075 mol<\/p>\n\n\n\n

    \n\n\n\n

    3. Calculate mass of fluorescein:<\/strong><\/h3>\n\n\n\n

    Given:<\/p>\n\n\n\n

      \n
    • MM of fluorescein = 332.21 g\/mol<\/li>\n<\/ul>\n\n\n\n

      Mass of fluorescein=3.03\u00d710\u22125 mol\u00d7332.21 g\/mol=0.01006 g\\text{Mass of fluorescein} = 3.03 \\times 10^{-5} \\text{ mol} \\times 332.21 \\text{ g\/mol} = 0.01006 \\text{ g}<\/p>\n\n\n\n

      \n\n\n\n

      4. Calculate concentration in ppm:<\/strong><\/h3>\n\n\n\n

      ppm is defined as mg of solute per liter (mg\/L)<\/strong> of solution.<\/p>\n\n\n\n

        \n
      • Final volume of stock solution = 100.00 mL = 0.10000 L<\/li>\n\n\n\n
      • Convert 0.01006 g to mg:<\/li>\n<\/ul>\n\n\n\n

        0.01006 g=10.06 mg0.01006 \\text{ g} = 10.06 \\text{ mg}<\/p>\n\n\n\n

        Now: ppm=10.06 mg0.10000 L=100.6 ppm\\text{ppm} = \\frac{10.06 \\text{ mg}}{0.10000 \\text{ L}} = 100.6 \\text{ ppm}<\/p>\n\n\n\n

        \n\n\n\n

        Summary:<\/strong><\/h3>\n\n\n\n
          \n
        • The disodium fluorescein salt was converted to pure fluorescein after dissolving.<\/li>\n\n\n\n
        • Moles were calculated using the salt\u2019s molar mass.<\/li>\n\n\n\n
        • Fluorescein mass was then calculated using its own molar mass.<\/li>\n\n\n\n
        • The final step converted that mass to ppm based on the 100 mL total solution volume.<\/li>\n<\/ul>\n\n\n\n

          This approach properly accounts for the dissociation of the salt and uses the molar mass of fluorescein<\/strong> (not the salt) in the final ppm concentration calculation.<\/p>\n\n\n\n

          \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

          The technician began his analysis by preparing a stock solution of fluorescein. To prepare the fluorescein stock solution, 0.01140 g of disodium fluorescein salt (MM = 376.28 g\/mol) was weighed on the microanalytical balance. The salt was dissolved in 50 mL of 0.1 M NaOH and then analytically transferred to a 100.00 mL volumetric flask. […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-222981","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222981","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=222981"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222981\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=222981"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=222981"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=222981"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}