{"id":231172,"date":"2025-06-10T15:01:45","date_gmt":"2025-06-10T15:01:45","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=231172"},"modified":"2025-06-10T15:01:47","modified_gmt":"2025-06-10T15:01:47","slug":"what-is-the-molar-mass-of-anhydrous-nickelii-chloride-hexahydrate-nicl-6h","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/10\/what-is-the-molar-mass-of-anhydrous-nickelii-chloride-hexahydrate-nicl-6h\/","title":{"rendered":"What is the molar mass of anhydrous nickel(II) chloride hexahydrate, NiCl.6H"},"content":{"rendered":"\n

What is the molar mass of anhydrous nickel(II) chloride hexahydrate, NiCl
.6H
O? 237.69 g\/mol 129.60 g\/mol 21411.12 g\/mol 147.62 g\/mol 108.10 g\/mol<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Determining Molar Mass<\/strong><\/p>\n\n\n\n

I’ve started by deconstructing the user’s request, focusing on the core question of determining the molar mass of anhydrous nickel(II) chloride hexahydrate. I’m now identifying the key components \u2013 Nickel(II) Chloride and its six water molecules of hydration. I’ll need to calculate each component’s molar mass and combine them accurately.<\/p>\n\n\n\n

Refining Calculation Approach<\/strong><\/p>\n\n\n\n

I’ve moved on to the practical calculation phase. The formula NiCl\u2082\u00b76H\u2082O requires me to determine the atomic weights of nickel, chlorine, hydrogen, and oxygen. Then, I must multiply the atomic weight of each element by its subscript within the formula. Finally, I’ll need to sum those individual results to find the total molar mass of the compound.<\/p>\n\n\n\n

Calculating Each Element’s Mass<\/strong><\/p>\n\n\n\n

I’m now calculating the molar mass using the formula NiCl\u2082\u00b76H\u2082O. I’ve looked up the atomic masses and determined the number of each atom present in the compound. I’ll multiply the atomic mass by the number of atoms for each element and sum those values. The goal is now to arrive at the correct molar mass within the answer choices provided.<\/p>\n\n\n\n

Finalizing Molar Mass Calculation<\/strong><\/p>\n\n\n\n

I’ve carefully calculated the molar mass, totaling each element’s mass contribution. Nickel, chlorine, hydrogen, and oxygen contributions were summed. The calculated value aligns closely with answer choice A, 237.69 g\/mol , the molar mass of the hexahydrate. The calculated value of 129.59 g\/mol , matches the choice B, and it is the molar mass of anhydrous nickel(II) chloride. The question’s wording was deliberately confusing.<\/p>\n\n\n\n

Calculating the Correct Option<\/strong><\/p>\n\n\n\n

I’ve carefully calculated the molar mass of the hexahydrate, NiCl\u2082\u00b76H\u2082O. My initial calculation yielded approximately 237.7 g\/mol . A re-run of the calculation, using precise atomic masses, yielded 237.689 g\/mol , which is very close to 237.69 g\/mol , choice A. I’ve now confirmed that the slight difference is likely due to the precision of the atomic masses used. The molar mass of the anhydrous form, which matches choice B, is indeed 129.59 g\/mol . This confirms the question intends me to calculate the hexahydrate, ignoring the word ‘anhydrous’.<\/p>\n\n\n\n

Revisiting the Question’s Wording<\/strong><\/p>\n\n\n\n

I’ve re-read the question carefully and noticed the “anhydrous” descriptor conflicts with the compound formula. It’s clear I need to focus on calculating the molar mass of the hexahydrate despite the misleading word. I’m now making sure the explanation emphasizes this crucial point.<\/p>\n\n\n\n

Analyzing the Formula<\/strong><\/p>\n\n\n\n

I’m now carefully analyzing the chemical formula. The formula NiCl\u2082\u00b76H\u2082O indicates that the calculation of molar mass must include both the nickel(II) chloride component and the water of hydration. I’ll need to determine the molar mass of each part, followed by their sum.<\/p>\n\n\n\n

Collapse to hide model thoughtschevron_righteditmore_vert<\/p>\n\n\n\n

The correct answer is 237.69 g\/mol <\/strong>.<\/p>\n\n\n\n

Explanation<\/h3>\n\n\n\n

The molar mass of a compound is the sum of the atomic masses of all atoms present in one mole of its formula unit. The chemical formula provided is NiCl\u2082\u00b76H\u2082O, which represents nickel(II) chloride hexahydrate. The term “anhydrous” in the question appears to be an error, as it means “without water,” which contradicts both the term “hexahydrate” and the given formula. The calculation must proceed based on the formula NiCl\u2082\u00b76H\u2082O.<\/p>\n\n\n\n

This formula indicates that for every one formula unit of nickel(II) chloride (NiCl\u2082), there are six molecules of water (H\u2082O) of hydration. The central dot (\u00b7) signifies that these water molecules are chemically bound within the crystal structure of the salt.<\/p>\n\n\n\n

To calculate the molar mass, one must determine the total mass contribution of each element in the formula.<\/p>\n\n\n\n

    \n
  1. Identify the elements and their counts:<\/strong>\n
      \n
    • Nickel (Ni): 1 atom<\/li>\n\n\n\n
    • Chlorine (Cl): 2 atoms<\/li>\n\n\n\n
    • Hydrogen (H): 6 water molecules \u00d7 2 hydrogen atoms\/molecule = 12 atoms<\/li>\n\n\n\n
    • Oxygen (O): 6 water molecules \u00d7 1 oxygen atom\/molecule = 6 atoms<\/li>\n<\/ul>\n<\/li>\n\n\n\n
    • Use the standard atomic masses for each element:<\/strong>\n
        \n
      • Atomic mass of Ni \u2248 58.69\u00a0g\/mol<\/li>\n\n\n\n
      • Atomic mass of Cl \u2248 35.45\u00a0g\/mol<\/li>\n\n\n\n
      • Atomic mass of H \u2248 1.01\u00a0g\/mol<\/li>\n\n\n\n
      • Atomic mass of O \u2248 16.00\u00a0g\/mol<\/li>\n<\/ul>\n<\/li>\n\n\n\n
      • Calculate the total mass for each element in the compound:<\/strong>\n
          \n
        • Mass of Ni: 1 \u00d7 58.69\u00a0g\/mol\u00a0= 58.69\u00a0g\/mol<\/li>\n\n\n\n
        • Mass of Cl: 2 \u00d7 35.45\u00a0g\/mol\u00a0= 70.90\u00a0g\/mol<\/li>\n\n\n\n
        • Mass of H: 12 \u00d7 1.01\u00a0g\/mol\u00a0= 12.12\u00a0g\/mol<\/li>\n\n\n\n
        • Mass of O: 6 \u00d7 16.00\u00a0g\/mol\u00a0= 96.00\u00a0g\/mol<\/li>\n<\/ul>\n<\/li>\n\n\n\n
        • Sum the individual masses to find the total molar mass:<\/strong>
          Molar Mass = 58.69 + 70.90 + 12.12 + 96.00 =\u00a0237.71\u00a0g\/mol<\/strong><\/li>\n<\/ol>\n\n\n\n

          This value is closest to the option 237.69 g\/mol <\/strong>, with the minor difference arising from rounding of atomic masses. The option 129.60 g\/mol corresponds to the anhydrous form, NiCl\u2082, which ignores the six water molecules.<\/p>\n\n\n\n

          \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

          What is the molar mass of anhydrous nickel(II) chloride hexahydrate, NiCl.6HO? 237.69 g\/mol 129.60 g\/mol 21411.12 g\/mol 147.62 g\/mol 108.10 g\/mol The Correct Answer and Explanation is: Determining Molar Mass I’ve started by deconstructing the user’s request, focusing on the core question of determining the molar mass of anhydrous nickel(II) chloride hexahydrate. I’m now identifying […]<\/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-231172","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231172","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=231172"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231172\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=231172"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=231172"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=231172"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}