{"id":183632,"date":"2025-01-16T18:31:59","date_gmt":"2025-01-16T18:31:59","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=183632"},"modified":"2025-01-16T18:32:01","modified_gmt":"2025-01-16T18:32:01","slug":"the-density-of-water-h201-is-1-0-g-ml","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/16\/the-density-of-water-h201-is-1-0-g-ml\/","title":{"rendered":"The density of water, H20(1), is 1.0 g\/mL"},"content":{"rendered":"\n<p>The density of water, H20(1), is 1.0 g\/mL. How many atoms of oxygen are present in 2.5 L of pure water? A) 8.4 x 10^25 B) 1.4 x 10^21 C) 2.3 x 10^-22 D) 1.5 x 10^27 E) 2.7 x 10^26 The mass of 2.5 L of water is mass = dV = 1.00 g mL^-1 x 2500 mL = 2500 g Note the conversion from L to mL (1 L = 1000 mL). The amount of water is thus2500 g Moles of H_2O= m_H2O\/M_H2O =2500 g\/ 18.015 g mol^-1=13_8 molMultiplying by the Avogadro constant gives the number of water molecules which equals the number of oxygen atoms since there is one oxygen atom per water molecule. number of O atoms = 13_8 x 6.022&#215;1023 atoms ml^-1 = 8.4&#215;10^25Shouldn&#8217;t it be 2H2 + O2 -> 2H2O? but how come in the picture i&#8217;ve attached, it is 1 oxygen = 1 h2o<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>Let&#8217;s walk through the calculation and explanation step by step, addressing the process and your question:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Given Data:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Density of water (H\u2082O) = 1.0 g\/mL<\/li>\n\n\n\n<li>Volume of water = 2.5 L = 2500 mL<\/li>\n\n\n\n<li>Molar mass of H\u2082O = 18.015 g\/mol<\/li>\n\n\n\n<li>Avogadro&#8217;s number (NA) = 6.022 \u00d7 10\u00b2\u00b3 atoms\/mol<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate the Mass of Water<\/h3>\n\n\n\n<p>From the given density of water (1.0 g\/mL), we can calculate the mass of the 2.5 L of water:<br>[<br>\\text{Mass} = \\text{Density} \\times \\text{Volume} = 1.0 \\, \\text{g\/mL} \\times 2500 \\, \\text{mL} = 2500 \\, \\text{g}<br>]<br>Thus, the mass of 2.5 L of water is 2500 grams.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Calculate the Number of Moles of H\u2082O<\/h3>\n\n\n\n<p>Now, let&#8217;s determine how many moles of H\u2082O are present in 2500 grams. Using the molar mass of water (18.015 g\/mol):<br>[<br>\\text{Moles of H\u2082O} = \\frac{\\text{Mass of H\u2082O}}{\\text{Molar mass of H\u2082O}} = \\frac{2500 \\, \\text{g}}{18.015 \\, \\text{g\/mol}} \\approx 138.6 \\, \\text{mol}<br>]<br>This means there are approximately 138.6 moles of water molecules in 2.5 L of water.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate the Number of Water Molecules<\/h3>\n\n\n\n<p>Next, we use Avogadro&#8217;s number to determine the number of water molecules:<br>[<br>\\text{Number of H\u2082O molecules} = 138.6 \\, \\text{mol} \\times 6.022 \\times 10^{23} \\, \\text{molecules\/mol} \\approx 8.36 \\times 10^{25} \\, \\text{molecules}<br>]<br>Thus, the number of water molecules in 2.5 L of water is approximately (8.4 \\times 10^{25}).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Number of Oxygen Atoms<\/h3>\n\n\n\n<p>Since each water molecule (H\u2082O) contains exactly one oxygen atom, the number of oxygen atoms in the 2.5 L of water will be the same as the number of water molecules. Therefore, the number of oxygen atoms is also:<br>[<br>\\text{Number of oxygen atoms} = 8.4 \\times 10^{25}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Answer:<\/h3>\n\n\n\n<p>The correct answer is <strong>A) (8.4 \\times 10^{25})<\/strong> oxygen atoms.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Why Is It 1 Oxygen Atom per Water Molecule?<\/h3>\n\n\n\n<p>In the structure of a water molecule (H\u2082O), there is one oxygen atom bonded to two hydrogen atoms. The balanced chemical equation for the formation of water from hydrogen and oxygen (2H\u2082 + O\u2082 \u2192 2H\u2082O) shows that one molecule of O\u2082 is required to form two water molecules. However, each individual water molecule (H\u2082O) contains exactly one oxygen atom, not two. So, despite oxygen being involved in the reaction with hydrogen to form water, each H\u2082O molecule still contains only one oxygen atom.<\/p>\n\n\n\n<p>Thus, for every mole of water, there is exactly one mole of oxygen atoms.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The density of water, H20(1), is 1.0 g\/mL. How many atoms of oxygen are present in 2.5 L of pure water? A) 8.4 x 10^25 B) 1.4 x 10^21 C) 2.3 x 10^-22 D) 1.5 x 10^27 E) 2.7 x 10^26 The mass of 2.5 L of water is mass = dV = 1.00 g [&hellip;]<\/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-183632","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183632","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=183632"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183632\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183632"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183632"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183632"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}