The concentration of lead in an industrial waste stream is 0.28 ppm. What is its molar concentration?<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the molar concentration<\/strong> (mol\/L) of lead (Pb) from a given concentration in ppm (parts per million)<\/strong>, we follow these steps:<\/p>\n\n\n\n Since 1 ppm \u2248 1 mg\/L (in dilute aqueous solutions like water):<\/p>\n\n\n\n [ [ [ [ The problem involves converting a concentration from ppm<\/strong> to molarity<\/strong> (mol\/L), which is essential in chemistry for understanding how many moles of solute<\/strong> are dissolved in a liter of solution<\/strong>.<\/p>\n\n\n\n Parts per million (ppm) is commonly used to express very dilute concentrations of substances, especially in environmental chemistry, such as contaminants in water or air. In aqueous solutions, 1 ppm is approximately equal to 1 mg of solute per liter of solution<\/strong>, assuming the density of water is about 1 g\/mL, which holds true for dilute solutions.<\/p>\n\n\n\n In this case, 0.28 ppm of lead means that there are 0.28 mg of lead in every liter of the waste stream. To work with molarity, which is based on grams per mole<\/strong>, we convert this to grams first:<\/p>\n\n\n\n [ Next, we convert grams to moles using the molar mass<\/strong> of lead. The molar mass (atomic weight) of lead is 207.2 g\/mol<\/strong>, meaning 1 mole of lead weighs 207.2 grams. By dividing the grams of lead per liter by its molar mass, we obtain its molar concentration:<\/p>\n\n\n\n [ This result represents a very low molar concentration<\/strong>, which is expected for a value expressed in parts per million. Environmental regulations often use such low concentrations to assess pollutant levels in water.<\/p>\n\n\n\n This method can be applied to other elements or compounds by simply adjusting the molar mass in the calculation.<\/p>\n","protected":false},"excerpt":{"rendered":" The concentration of lead in an industrial waste stream is 0.28 ppm. What is its molar concentration? The correct answer and explanation is : To calculate the molar concentration (mol\/L) of lead (Pb) from a given concentration in ppm (parts per million), we follow these steps: Given: Step 1: Convert ppm to mg\/L Since 1 […]<\/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-208004","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/208004","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=208004"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/208004\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=208004"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=208004"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=208004"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nGiven:<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\nStep 1: Convert ppm to mg\/L<\/strong><\/h3>\n\n\n\n
0.28 \\text{ ppm} = 0.28 \\text{ mg\/L}
]<\/p>\n\n\n\n
\n\n\n\nStep 2: Convert mg\/L to g\/L<\/strong><\/h3>\n\n\n\n
0.28 \\text{ mg\/L} = 0.00028 \\text{ g\/L}
]<\/p>\n\n\n\n
\n\n\n\nStep 3: Convert g\/L to mol\/L<\/strong><\/h3>\n\n\n\n
\\text{Molar concentration} = \\frac{\\text{grams per liter}}{\\text{molar mass}} = \\frac{0.00028}{207.2} \\approx 1.351 \\times 10^{-6} \\text{ mol\/L}
]<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answer:<\/strong><\/h3>\n\n\n\n
\\boxed{1.35 \\times 10^{-6} \\text{ mol\/L}}
]<\/p>\n\n\n\n
\n\n\n\n\ud83d\udcda Explanation (300+ words):<\/strong><\/h3>\n\n\n\n
0.28 \\text{ mg} = 0.00028 \\text{ g}
]<\/p>\n\n\n\n
\\frac{0.00028 \\text{ g}}{207.2 \\text{ g\/mol}} \\approx 1.35 \\times 10^{-6} \\text{ mol\/L}
]<\/p>\n\n\n\n