{"id":87667,"date":"2025-06-01T06:50:53","date_gmt":"2025-06-01T06:50:53","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87667"},"modified":"2025-06-01T06:50:53","modified_gmt":"2025-06-01T06:50:53","slug":"the-equivalent-weight-of-oxalic-acid-in-c%e2%82%82h%e2%82%82o%e2%82%84-2h%e2%82%82o-is","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-equivalent-weight-of-oxalic-acid-in-c%e2%82%82h%e2%82%82o%e2%82%84-2h%e2%82%82o-is\/","title":{"rendered":"The equivalent weight of oxalic acid in C\u2082H\u2082O\u2084.2H\u2082O is"},"content":{"rendered":"

The equivalent weight of oxalic acid in C\u2082H\u2082O\u2084.2H\u2082O is<\/p>\n

[amp_mcq option1=”45″ option2=”63″ option3=”90″ option4=”126″ correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-1 – 2019<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe formula given is for hydrated oxalic acid, C\u2082H\u2082O\u2084.2H\u2082O. Oxalic acid (C\u2082H\u2082O\u2084 or HOOC-COOH) is a diprotic acid, meaning it can donate two protons (H\u207a). The equivalent weight of an acid is its molar mass divided by the number of acidic hydrogens that participate in the reaction (also called its basicity). In typical acid-base reactions, oxalic acid dihydrate reacts as a diprotic acid, donating both protons.
\n<\/section>\n
\n– The molar mass of C\u2082H\u2082O\u2084 = (2 \u00d7 12.01) + (2 \u00d7 1.01) + (4 \u00d7 16.00) = 24.02 + 2.02 + 64.00 = 90.04 g\/mol.
\n– The molar mass of 2H\u2082O = 2 \u00d7 (2 \u00d7 1.01 + 16.00) = 2 \u00d7 18.02 = 36.04 g\/mol.
\n– The molar mass of C\u2082H\u2082O\u2084.2H\u2082O = 90.04 + 36.04 = 126.08 g\/mol.
\n– Oxalic acid is a diprotic acid, meaning it has 2 acidic hydrogens.
\n– Equivalent weight = Molar mass \/ number of acidic hydrogens = 126.08 g\/mol \/ 2 = 63.04 g\/equivalent.
\n– Among the options, 63 is the closest value.
\n<\/section>\n
\nIf the question were asking for the equivalent weight of *anhydrous* oxalic acid (C\u2082H\u2082O\u2084) reacting as a diprotic acid, the answer would be 90\/2 = 45. If it reacted as a monoprotic acid, the equivalent weight would be 90 or 126 depending on whether anhydrous or hydrated form is considered as the basis for molar mass. However, the most common standard use of oxalic acid in volumetric analysis is as the dihydrate, reacting as a diprotic acid.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

The equivalent weight of oxalic acid in C\u2082H\u2082O\u2084.2H\u2082O is [amp_mcq option1=”45″ option2=”63″ option3=”90″ option4=”126″ correct=”option2″] This question was previously asked in UPSC NDA-1 – 2019 Download PDFAttempt Online The formula given is for hydrated oxalic acid, C\u2082H\u2082O\u2084.2H\u2082O. Oxalic acid (C\u2082H\u2082O\u2084 or HOOC-COOH) is a diprotic acid, meaning it can donate two protons (H\u207a). The equivalent … <\/p>\n

Detailed SolutionThe equivalent weight of oxalic acid in C\u2082H\u2082O\u2084.2H\u2082O is<\/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":[1119,1202,1096],"class_list":["post-87667","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1119","tag-acid-base-and-salt","tag-chemistry","no-featured-image-padding"],"yoast_head":"\nThe equivalent weight of oxalic acid in C\u2082H\u2082O\u2084.2H\u2082O is<\/title>\n<meta name=\"description\" content=\"The formula given is for hydrated oxalic acid, C\u2082H\u2082O\u2084.2H\u2082O. Oxalic acid (C\u2082H\u2082O\u2084 or HOOC-COOH) is a diprotic acid, meaning it can donate two protons (H\u207a). The equivalent weight of an acid is its molar mass divided by the number of acidic hydrogens that participate in the reaction (also called its basicity). In typical acid-base reactions, oxalic acid dihydrate reacts as a diprotic acid, donating both protons. - The molar mass of C\u2082H\u2082O\u2084 = (2 \u00d7 12.01) + (2 \u00d7 1.01) + (4 \u00d7 16.00) = 24.02 + 2.02 + 64.00 = 90.04 g\/mol. - The molar mass of 2H\u2082O = 2 \u00d7 (2 \u00d7 1.01 + 16.00) = 2 \u00d7 18.02 = 36.04 g\/mol. - The molar mass of C\u2082H\u2082O\u2084.2H\u2082O = 90.04 + 36.04 = 126.08 g\/mol. - Oxalic acid is a diprotic acid, meaning it has 2 acidic hydrogens. - Equivalent weight = Molar mass \/ number of acidic hydrogens = 126.08 g\/mol \/ 2 = 63.04 g\/equivalent. - Among the options, 63 is the closest value.\" \/>\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-equivalent-weight-of-oxalic-acid-in-c\u2082h\u2082o\u2084-2h\u2082o-is\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"The equivalent weight of oxalic acid in C\u2082H\u2082O\u2084.2H\u2082O is\" \/>\n<meta property=\"og:description\" content=\"The formula given is for hydrated oxalic acid, C\u2082H\u2082O\u2084.2H\u2082O. Oxalic acid (C\u2082H\u2082O\u2084 or HOOC-COOH) is a diprotic acid, meaning it can donate two protons (H\u207a). The equivalent weight of an acid is its molar mass divided by the number of acidic hydrogens that participate in the reaction (also called its basicity). In typical acid-base reactions, oxalic acid dihydrate reacts as a diprotic acid, donating both protons. - The molar mass of C\u2082H\u2082O\u2084 = (2 \u00d7 12.01) + (2 \u00d7 1.01) + (4 \u00d7 16.00) = 24.02 + 2.02 + 64.00 = 90.04 g\/mol. - The molar mass of 2H\u2082O = 2 \u00d7 (2 \u00d7 1.01 + 16.00) = 2 \u00d7 18.02 = 36.04 g\/mol. - The molar mass of C\u2082H\u2082O\u2084.2H\u2082O = 90.04 + 36.04 = 126.08 g\/mol. - Oxalic acid is a diprotic acid, meaning it has 2 acidic hydrogens. - Equivalent weight = Molar mass \/ number of acidic hydrogens = 126.08 g\/mol \/ 2 = 63.04 g\/equivalent. - Among the options, 63 is the closest value.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/the-equivalent-weight-of-oxalic-acid-in-c\u2082h\u2082o\u2084-2h\u2082o-is\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T06:50:53+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 equivalent weight of oxalic acid in C\u2082H\u2082O\u2084.2H\u2082O is","description":"The formula given is for hydrated oxalic acid, C\u2082H\u2082O\u2084.2H\u2082O. Oxalic acid (C\u2082H\u2082O\u2084 or HOOC-COOH) is a diprotic acid, meaning it can donate two protons (H\u207a). The equivalent weight of an acid is its molar mass divided by the number of acidic hydrogens that participate in the reaction (also called its basicity). In typical acid-base reactions, oxalic acid dihydrate reacts as a diprotic acid, donating both protons. - The molar mass of C\u2082H\u2082O\u2084 = (2 \u00d7 12.01) + (2 \u00d7 1.01) + (4 \u00d7 16.00) = 24.02 + 2.02 + 64.00 = 90.04 g\/mol. - The molar mass of 2H\u2082O = 2 \u00d7 (2 \u00d7 1.01 + 16.00) = 2 \u00d7 18.02 = 36.04 g\/mol. - The molar mass of C\u2082H\u2082O\u2084.2H\u2082O = 90.04 + 36.04 = 126.08 g\/mol. - Oxalic acid is a diprotic acid, meaning it has 2 acidic hydrogens. - Equivalent weight = Molar mass \/ number of acidic hydrogens = 126.08 g\/mol \/ 2 = 63.04 g\/equivalent. - Among the options, 63 is the closest value.","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-equivalent-weight-of-oxalic-acid-in-c\u2082h\u2082o\u2084-2h\u2082o-is\/","og_locale":"en_US","og_type":"article","og_title":"The equivalent weight of oxalic acid in C\u2082H\u2082O\u2084.2H\u2082O is","og_description":"The formula given is for hydrated oxalic acid, C\u2082H\u2082O\u2084.2H\u2082O. Oxalic acid (C\u2082H\u2082O\u2084 or HOOC-COOH) is a diprotic acid, meaning it can donate two protons (H\u207a). The equivalent weight of an acid is its molar mass divided by the number of acidic hydrogens that participate in the reaction (also called its basicity). In typical acid-base reactions, oxalic acid dihydrate reacts as a diprotic acid, donating both protons. - The molar mass of C\u2082H\u2082O\u2084 = (2 \u00d7 12.01) + (2 \u00d7 1.01) + (4 \u00d7 16.00) = 24.02 + 2.02 + 64.00 = 90.04 g\/mol. - The molar mass of 2H\u2082O = 2 \u00d7 (2 \u00d7 1.01 + 16.00) = 2 \u00d7 18.02 = 36.04 g\/mol. - The molar mass of C\u2082H\u2082O\u2084.2H\u2082O = 90.04 + 36.04 = 126.08 g\/mol. - Oxalic acid is a diprotic acid, meaning it has 2 acidic hydrogens. - Equivalent weight = Molar mass \/ number of acidic hydrogens = 126.08 g\/mol \/ 2 = 63.04 g\/equivalent. - Among the options, 63 is the closest value.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/the-equivalent-weight-of-oxalic-acid-in-c\u2082h\u2082o\u2084-2h\u2082o-is\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T06:50:53+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-equivalent-weight-of-oxalic-acid-in-c%e2%82%82h%e2%82%82o%e2%82%84-2h%e2%82%82o-is\/","url":"https:\/\/exam.pscnotes.com\/mcq\/the-equivalent-weight-of-oxalic-acid-in-c%e2%82%82h%e2%82%82o%e2%82%84-2h%e2%82%82o-is\/","name":"The equivalent weight of oxalic acid in C\u2082H\u2082O\u2084.2H\u2082O is","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T06:50:53+00:00","dateModified":"2025-06-01T06:50:53+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The formula given is for hydrated oxalic acid, C\u2082H\u2082O\u2084.2H\u2082O. Oxalic acid (C\u2082H\u2082O\u2084 or HOOC-COOH) is a diprotic acid, meaning it can donate two protons (H\u207a). The equivalent weight of an acid is its molar mass divided by the number of acidic hydrogens that participate in the reaction (also called its basicity). In typical acid-base reactions, oxalic acid dihydrate reacts as a diprotic acid, donating both protons. - The molar mass of C\u2082H\u2082O\u2084 = (2 \u00d7 12.01) + (2 \u00d7 1.01) + (4 \u00d7 16.00) = 24.02 + 2.02 + 64.00 = 90.04 g\/mol. - The molar mass of 2H\u2082O = 2 \u00d7 (2 \u00d7 1.01 + 16.00) = 2 \u00d7 18.02 = 36.04 g\/mol. - The molar mass of C\u2082H\u2082O\u2084.2H\u2082O = 90.04 + 36.04 = 126.08 g\/mol. - Oxalic acid is a diprotic acid, meaning it has 2 acidic hydrogens. - Equivalent weight = Molar mass \/ number of acidic hydrogens = 126.08 g\/mol \/ 2 = 63.04 g\/equivalent. - Among the options, 63 is the closest value.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/the-equivalent-weight-of-oxalic-acid-in-c%e2%82%82h%e2%82%82o%e2%82%84-2h%e2%82%82o-is\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/the-equivalent-weight-of-oxalic-acid-in-c%e2%82%82h%e2%82%82o%e2%82%84-2h%e2%82%82o-is\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/the-equivalent-weight-of-oxalic-acid-in-c%e2%82%82h%e2%82%82o%e2%82%84-2h%e2%82%82o-is\/#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 equivalent weight of oxalic acid in C\u2082H\u2082O\u2084.2H\u2082O is"}]},{"@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\/87667","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=87667"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87667\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=87667"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=87667"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=87667"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}