{"id":88049,"date":"2025-06-01T07:00:26","date_gmt":"2025-06-01T07:00:26","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88049"},"modified":"2025-06-01T07:00:26","modified_gmt":"2025-06-01T07:00:26","slug":"the-relative-atomic-mass-of-boron-which-exists-in-two-isotopic-forms","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-relative-atomic-mass-of-boron-which-exists-in-two-isotopic-forms\/","title":{"rendered":"The relative atomic mass of boron (which exists in two isotopic forms"},"content":{"rendered":"

The relative atomic mass of boron (which exists in two isotopic forms 10<\/sup>B and 11<\/sup>B) is 10\u00b781. What will be the abundance of 10<\/sup>B and 11<\/sup>B, respectively (consider a sample of 100 atoms) ?<\/p>\n

[amp_mcq option1=”19% and 81%” option2=”81% and 19%” option3=”38% and 62%” option4=”62% and 38%” correct=”option1″]<\/p>\n

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This question was previously asked in<\/div>\n
UPSC NDA-1 – 2024<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nLet the abundance of 10<\/sup>B be x and the abundance of 11<\/sup>B be y. The total abundance is 100%, so x + y = 1 (or 100 if using percentages). The average atomic mass is calculated as the weighted average of the isotopic masses: (mass of 10<\/sup>B * x) + (mass of 11<\/sup>B * y) = Average Atomic Mass. Using approximate integer masses (10 for 10<\/sup>B and 11 for 11<\/sup>B), we have: 10x + 11y = 10.81. Since x + y = 1, we can write y = 1 – x. Substituting this into the equation: 10x + 11(1 – x) = 10.81 => 10x + 11 – 11x = 10.81 => 11 – x = 10.81 => x = 11 – 10.81 = 0.19. So, the abundance of 10<\/sup>B is 0.19 or 19%. Then, y = 1 – x = 1 – 0.19 = 0.81 or 81%. The abundance of 10<\/sup>B and 11<\/sup>B, respectively, is 19% and 81%.
\n<\/section>\n
\nThe average atomic mass of an element is the weighted average of the masses of its isotopes, where the weights are their natural abundances. This principle is used to determine the relative abundance of isotopes if the average atomic mass and isotopic masses are known.
\n<\/section>\n
\nIsotopes are atoms of the same element that have the same number of protons but different numbers of neutrons, resulting in different atomic masses. Boron’s two stable isotopes are 10<\/sup>B (with 5 protons and 5 neutrons) and 11<\/sup>B (with 5 protons and 6 neutrons). The natural abundance of isotopes is relatively constant for a given element across the Earth.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

The relative atomic mass of boron (which exists in two isotopic forms 10B and 11B) is 10\u00b781. What will be the abundance of 10B and 11B, respectively (consider a sample of 100 atoms) ? [amp_mcq option1=”19% and 81%” option2=”81% and 19%” option3=”38% and 62%” option4=”62% and 38%” correct=”option1″] This question was previously asked in UPSC … <\/p>\n

Detailed SolutionThe relative atomic mass of boron (which exists in two isotopic forms<\/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":[1093],"tags":[1103,1162,1096],"class_list":["post-88049","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1103","tag-atomic-structure","tag-chemistry","no-featured-image-padding"],"yoast_head":"\nThe relative atomic mass of boron (which exists in two isotopic forms<\/title>\n<meta name=\"description\" content=\"Let the abundance of 10B be x and the abundance of 11B be y. The total abundance is 100%, so x + y = 1 (or 100 if using percentages). The average atomic mass is calculated as the weighted average of the isotopic masses: (mass of 10B * x) + (mass of 11B * y) = Average Atomic Mass. Using approximate integer masses (10 for 10B and 11 for 11B), we have: 10x + 11y = 10.81. Since x + y = 1, we can write y = 1 - x. Substituting this into the equation: 10x + 11(1 - x) = 10.81 => 10x + 11 - 11x = 10.81 => 11 - x = 10.81 => x = 11 - 10.81 = 0.19. So, the abundance of 10B is 0.19 or 19%. Then, y = 1 - x = 1 - 0.19 = 0.81 or 81%. The abundance of 10B and 11B, respectively, is 19% and 81%. The average atomic mass of an element is the weighted average of the masses of its isotopes, where the weights are their natural abundances. This principle is used to determine the relative abundance of isotopes if the average atomic mass and isotopic masses are known.\" \/>\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\/the-relative-atomic-mass-of-boron-which-exists-in-two-isotopic-forms\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"The relative atomic mass of boron (which exists in two isotopic forms\" \/>\n<meta property=\"og:description\" content=\"Let the abundance of 10B be x and the abundance of 11B be y. The total abundance is 100%, so x + y = 1 (or 100 if using percentages). The average atomic mass is calculated as the weighted average of the isotopic masses: (mass of 10B * x) + (mass of 11B * y) = Average Atomic Mass. Using approximate integer masses (10 for 10B and 11 for 11B), we have: 10x + 11y = 10.81. Since x + y = 1, we can write y = 1 - x. Substituting this into the equation: 10x + 11(1 - x) = 10.81 => 10x + 11 - 11x = 10.81 => 11 - x = 10.81 => x = 11 - 10.81 = 0.19. So, the abundance of 10B is 0.19 or 19%. Then, y = 1 - x = 1 - 0.19 = 0.81 or 81%. The abundance of 10B and 11B, respectively, is 19% and 81%. The average atomic mass of an element is the weighted average of the masses of its isotopes, where the weights are their natural abundances. This principle is used to determine the relative abundance of isotopes if the average atomic mass and isotopic masses are known.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/the-relative-atomic-mass-of-boron-which-exists-in-two-isotopic-forms\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:00:26+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":"The relative atomic mass of boron (which exists in two isotopic forms","description":"Let the abundance of 10B be x and the abundance of 11B be y. The total abundance is 100%, so x + y = 1 (or 100 if using percentages). The average atomic mass is calculated as the weighted average of the isotopic masses: (mass of 10B * x) + (mass of 11B * y) = Average Atomic Mass. Using approximate integer masses (10 for 10B and 11 for 11B), we have: 10x + 11y = 10.81. Since x + y = 1, we can write y = 1 - x. Substituting this into the equation: 10x + 11(1 - x) = 10.81 => 10x + 11 - 11x = 10.81 => 11 - x = 10.81 => x = 11 - 10.81 = 0.19. So, the abundance of 10B is 0.19 or 19%. Then, y = 1 - x = 1 - 0.19 = 0.81 or 81%. The abundance of 10B and 11B, respectively, is 19% and 81%. The average atomic mass of an element is the weighted average of the masses of its isotopes, where the weights are their natural abundances. This principle is used to determine the relative abundance of isotopes if the average atomic mass and isotopic masses are known.","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\/the-relative-atomic-mass-of-boron-which-exists-in-two-isotopic-forms\/","og_locale":"en_US","og_type":"article","og_title":"The relative atomic mass of boron (which exists in two isotopic forms","og_description":"Let the abundance of 10B be x and the abundance of 11B be y. The total abundance is 100%, so x + y = 1 (or 100 if using percentages). The average atomic mass is calculated as the weighted average of the isotopic masses: (mass of 10B * x) + (mass of 11B * y) = Average Atomic Mass. Using approximate integer masses (10 for 10B and 11 for 11B), we have: 10x + 11y = 10.81. Since x + y = 1, we can write y = 1 - x. Substituting this into the equation: 10x + 11(1 - x) = 10.81 => 10x + 11 - 11x = 10.81 => 11 - x = 10.81 => x = 11 - 10.81 = 0.19. So, the abundance of 10B is 0.19 or 19%. Then, y = 1 - x = 1 - 0.19 = 0.81 or 81%. The abundance of 10B and 11B, respectively, is 19% and 81%. The average atomic mass of an element is the weighted average of the masses of its isotopes, where the weights are their natural abundances. This principle is used to determine the relative abundance of isotopes if the average atomic mass and isotopic masses are known.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/the-relative-atomic-mass-of-boron-which-exists-in-two-isotopic-forms\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:00:26+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\/the-relative-atomic-mass-of-boron-which-exists-in-two-isotopic-forms\/","url":"https:\/\/exam.pscnotes.com\/mcq\/the-relative-atomic-mass-of-boron-which-exists-in-two-isotopic-forms\/","name":"The relative atomic mass of boron (which exists in two isotopic forms","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:00:26+00:00","dateModified":"2025-06-01T07:00:26+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"Let the abundance of 10B be x and the abundance of 11B be y. The total abundance is 100%, so x + y = 1 (or 100 if using percentages). The average atomic mass is calculated as the weighted average of the isotopic masses: (mass of 10B * x) + (mass of 11B * y) = Average Atomic Mass. Using approximate integer masses (10 for 10B and 11 for 11B), we have: 10x + 11y = 10.81. Since x + y = 1, we can write y = 1 - x. Substituting this into the equation: 10x + 11(1 - x) = 10.81 => 10x + 11 - 11x = 10.81 => 11 - x = 10.81 => x = 11 - 10.81 = 0.19. So, the abundance of 10B is 0.19 or 19%. Then, y = 1 - x = 1 - 0.19 = 0.81 or 81%. The abundance of 10B and 11B, respectively, is 19% and 81%. The average atomic mass of an element is the weighted average of the masses of its isotopes, where the weights are their natural abundances. This principle is used to determine the relative abundance of isotopes if the average atomic mass and isotopic masses are known.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/the-relative-atomic-mass-of-boron-which-exists-in-two-isotopic-forms\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/the-relative-atomic-mass-of-boron-which-exists-in-two-isotopic-forms\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/the-relative-atomic-mass-of-boron-which-exists-in-two-isotopic-forms\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-1","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-1\/"},{"@type":"ListItem","position":3,"name":"The relative atomic mass of boron (which exists in two isotopic forms"}]},{"@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\/88049","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=88049"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88049\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=88049"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=88049"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=88049"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}