{"id":88799,"date":"2025-06-01T07:21:26","date_gmt":"2025-06-01T07:21:26","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88799"},"modified":"2025-06-01T07:21:26","modified_gmt":"2025-06-01T07:21:26","slug":"chlorine-occurs-in-nature-in-two-isotopic-forms-of-masses-35-u-and-37","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/chlorine-occurs-in-nature-in-two-isotopic-forms-of-masses-35-u-and-37\/","title":{"rendered":"Chlorine occurs in nature in two isotopic forms of masses 35 u and 37"},"content":{"rendered":"

Chlorine occurs in nature in two isotopic forms of masses 35 u and 37 u in the ratio of 3 : 1 respectively. What is the average atomic mass of the Chlorine atom ?<\/p>\n

[amp_mcq option1=”36\u00b71 u” option2=”35\u00b75 u” option3=”36\u00b75 u” option4=”35\u00b71 u” correct=”option2″]<\/p>\n

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This question was previously asked in<\/div>\n
UPSC NDA-2 – 2021<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
Chlorine exists as two isotopes: $^{35}$Cl with a mass of 35 u and $^{37}$Cl with a mass of 37 u. They are present in nature in a ratio of 3:1. This means that out of every 4 chlorine atoms, approximately 3 have a mass of 35 u and 1 has a mass of 37 u. The average atomic mass is calculated as a weighted average of the masses of the isotopes based on their relative abundance. The relative abundance of $^{35}$Cl is 3\/4 or 0.75, and the relative abundance of $^{37}$Cl is 1\/4 or 0.25.<\/section>\n
Average atomic mass = ($\\text{Mass of isotope 1} \\times \\text{Fractional abundance of isotope 1}$) + ($\\text{Mass of isotope 2} \\times \\text{Fractional abundance of isotope 2}$)
\nAverage atomic mass = $(35 \\text{ u} \\times 3\/4) + (37 \\text{ u} \\times 1\/4)$
\nAverage atomic mass = $(105\/4) \\text{ u} + (37\/4) \\text{ u}$
\nAverage atomic mass = $(105 + 37)\/4 \\text{ u}$
\nAverage atomic mass = $142\/4 \\text{ u}$
\nAverage atomic mass = $35.5 \\text{ u}$.<\/section>\n
Atomic mass unit (u) is a standard unit used for indicating atomic and molecular masses. Average atomic mass is the value listed for an element on the periodic table. It reflects the natural isotopic composition of the element.<\/section>\n","protected":false},"excerpt":{"rendered":"

Chlorine occurs in nature in two isotopic forms of masses 35 u and 37 u in the ratio of 3 : 1 respectively. What is the average atomic mass of the Chlorine atom ? [amp_mcq option1=”36\u00b71 u” option2=”35\u00b75 u” option3=”36\u00b75 u” option4=”35\u00b71 u” correct=”option2″] This question was previously asked in UPSC NDA-2 – 2021 Download … <\/p>\n

Detailed SolutionChlorine occurs in nature in two isotopic forms of masses 35 u and 37<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1094],"tags":[1110,1162,1096],"class_list":["post-88799","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1110","tag-atomic-structure","tag-chemistry","no-featured-image-padding"],"yoast_head":"\nChlorine occurs in nature in two isotopic forms of masses 35 u and 37<\/title>\n<meta name=\"description\" content=\"Chlorine exists as two isotopes: $^{35}$Cl with a mass of 35 u and $^{37}$Cl with a mass of 37 u. They are present in nature in a ratio of 3:1. This means that out of every 4 chlorine atoms, approximately 3 have a mass of 35 u and 1 has a mass of 37 u. The average atomic mass is calculated as a weighted average of the masses of the isotopes based on their relative abundance. The relative abundance of $^{35}$Cl is 3\/4 or 0.75, and the relative abundance of $^{37}$Cl is 1\/4 or 0.25. Average atomic mass = ($text{Mass of isotope 1} times text{Fractional abundance of isotope 1}$) + ($text{Mass of isotope 2} times text{Fractional abundance of isotope 2}$) Average atomic mass = $(35 text{ u} times 3\/4) + (37 text{ u} times 1\/4)$ Average atomic mass = $(105\/4) text{ u} + (37\/4) text{ u}$ Average atomic mass = $(105 + 37)\/4 text{ u}$ Average atomic mass = $142\/4 text{ u}$ Average atomic mass = $35.5 text{ u}$.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/chlorine-occurs-in-nature-in-two-isotopic-forms-of-masses-35-u-and-37\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Chlorine occurs in nature in two isotopic forms of masses 35 u and 37\" \/>\n<meta property=\"og:description\" content=\"Chlorine exists as two isotopes: $^{35}$Cl with a mass of 35 u and $^{37}$Cl with a mass of 37 u. They are present in nature in a ratio of 3:1. This means that out of every 4 chlorine atoms, approximately 3 have a mass of 35 u and 1 has a mass of 37 u. The average atomic mass is calculated as a weighted average of the masses of the isotopes based on their relative abundance. The relative abundance of $^{35}$Cl is 3\/4 or 0.75, and the relative abundance of $^{37}$Cl is 1\/4 or 0.25. Average atomic mass = ($text{Mass of isotope 1} times text{Fractional abundance of isotope 1}$) + ($text{Mass of isotope 2} times text{Fractional abundance of isotope 2}$) Average atomic mass = $(35 text{ u} times 3\/4) + (37 text{ u} times 1\/4)$ Average atomic mass = $(105\/4) text{ u} + (37\/4) text{ u}$ Average atomic mass = $(105 + 37)\/4 text{ u}$ Average atomic mass = $142\/4 text{ u}$ Average atomic mass = $35.5 text{ u}$.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/chlorine-occurs-in-nature-in-two-isotopic-forms-of-masses-35-u-and-37\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:21:26+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"Chlorine occurs in nature in two isotopic forms of masses 35 u and 37","description":"Chlorine exists as two isotopes: $^{35}$Cl with a mass of 35 u and $^{37}$Cl with a mass of 37 u. They are present in nature in a ratio of 3:1. This means that out of every 4 chlorine atoms, approximately 3 have a mass of 35 u and 1 has a mass of 37 u. The average atomic mass is calculated as a weighted average of the masses of the isotopes based on their relative abundance. The relative abundance of $^{35}$Cl is 3\/4 or 0.75, and the relative abundance of $^{37}$Cl is 1\/4 or 0.25. Average atomic mass = ($text{Mass of isotope 1} times text{Fractional abundance of isotope 1}$) + ($text{Mass of isotope 2} times text{Fractional abundance of isotope 2}$) Average atomic mass = $(35 text{ u} times 3\/4) + (37 text{ u} times 1\/4)$ Average atomic mass = $(105\/4) text{ u} + (37\/4) text{ u}$ Average atomic mass = $(105 + 37)\/4 text{ u}$ Average atomic mass = $142\/4 text{ u}$ Average atomic mass = $35.5 text{ u}$.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/chlorine-occurs-in-nature-in-two-isotopic-forms-of-masses-35-u-and-37\/","og_locale":"en_US","og_type":"article","og_title":"Chlorine occurs in nature in two isotopic forms of masses 35 u and 37","og_description":"Chlorine exists as two isotopes: $^{35}$Cl with a mass of 35 u and $^{37}$Cl with a mass of 37 u. They are present in nature in a ratio of 3:1. This means that out of every 4 chlorine atoms, approximately 3 have a mass of 35 u and 1 has a mass of 37 u. The average atomic mass is calculated as a weighted average of the masses of the isotopes based on their relative abundance. The relative abundance of $^{35}$Cl is 3\/4 or 0.75, and the relative abundance of $^{37}$Cl is 1\/4 or 0.25. Average atomic mass = ($text{Mass of isotope 1} times text{Fractional abundance of isotope 1}$) + ($text{Mass of isotope 2} times text{Fractional abundance of isotope 2}$) Average atomic mass = $(35 text{ u} times 3\/4) + (37 text{ u} times 1\/4)$ Average atomic mass = $(105\/4) text{ u} + (37\/4) text{ u}$ Average atomic mass = $(105 + 37)\/4 text{ u}$ Average atomic mass = $142\/4 text{ u}$ Average atomic mass = $35.5 text{ u}$.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/chlorine-occurs-in-nature-in-two-isotopic-forms-of-masses-35-u-and-37\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:21:26+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/chlorine-occurs-in-nature-in-two-isotopic-forms-of-masses-35-u-and-37\/","url":"https:\/\/exam.pscnotes.com\/mcq\/chlorine-occurs-in-nature-in-two-isotopic-forms-of-masses-35-u-and-37\/","name":"Chlorine occurs in nature in two isotopic forms of masses 35 u and 37","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:21:26+00:00","dateModified":"2025-06-01T07:21:26+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"Chlorine exists as two isotopes: $^{35}$Cl with a mass of 35 u and $^{37}$Cl with a mass of 37 u. They are present in nature in a ratio of 3:1. This means that out of every 4 chlorine atoms, approximately 3 have a mass of 35 u and 1 has a mass of 37 u. The average atomic mass is calculated as a weighted average of the masses of the isotopes based on their relative abundance. The relative abundance of $^{35}$Cl is 3\/4 or 0.75, and the relative abundance of $^{37}$Cl is 1\/4 or 0.25. Average atomic mass = ($\\text{Mass of isotope 1} \\times \\text{Fractional abundance of isotope 1}$) + ($\\text{Mass of isotope 2} \\times \\text{Fractional abundance of isotope 2}$) Average atomic mass = $(35 \\text{ u} \\times 3\/4) + (37 \\text{ u} \\times 1\/4)$ Average atomic mass = $(105\/4) \\text{ u} + (37\/4) \\text{ u}$ Average atomic mass = $(105 + 37)\/4 \\text{ u}$ Average atomic mass = $142\/4 \\text{ u}$ Average atomic mass = $35.5 \\text{ u}$.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/chlorine-occurs-in-nature-in-two-isotopic-forms-of-masses-35-u-and-37\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/chlorine-occurs-in-nature-in-two-isotopic-forms-of-masses-35-u-and-37\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/chlorine-occurs-in-nature-in-two-isotopic-forms-of-masses-35-u-and-37\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-2","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-2\/"},{"@type":"ListItem","position":3,"name":"Chlorine occurs in nature in two isotopic forms of masses 35 u and 37"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88799","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=88799"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88799\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=88799"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=88799"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=88799"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}