{"id":109,"date":"2015-11-30T13:13:00","date_gmt":"2015-11-30T13:13:00","guid":{"rendered":"https:\/\/rijankc.com.np\/2015\/11\/30\/08-molecular-weight\/"},"modified":"2024-11-03T22:07:01","modified_gmt":"2024-11-03T16:22:01","slug":"08-molecular-weight","status":"publish","type":"post","link":"https:\/\/rijankc.com.np\/?p=109","title":{"rendered":"08 Molecular Weight"},"content":{"rendered":"
\n
Molecular Weight is the number which shows how many times a molecule of a substance is heavier than 1\/12th<\/sup> part of an atom of C12<\/sup> isotope. In other words molecular weight is the sum of atomic weight of all atoms present in a molecule of a substance.
\n<\/a>eg. Mol. wt. of CaCO3<\/sub> =1\u00d740 + 1\u00d712 + 3\u00d716 = 100\u00a0<\/b><\/i>Gram Molecular Weight\u00a0<\/b><\/i>
\nWhen molecular weight of a substance is expressed in gram, then it is called gram molecular weight. One gram molecular weight equals to one mole and its contains 6.023\u00d71023<\/sup> molecules.
\nFor Eg.
\nMolecular Weight of CaCO3<\/sub> = 100
\n100gm of CaCO3<\/sub> = 1 gram molecular weight of CaCO3<\/sub>
\n=\u00a0 1 Mole of CaCO3<\/sub>
\n= 6.023\u00d71023<\/sup> molecules of CaCO3<\/sub><\/p>\n
<\/a><\/div>\n

Q. Calculate the number of moles and number of molecules present in 80gm of CaCO3<\/sub>.<\/b><\/i>
\nSolution:
\nMolecular Weight of CaCO3<\/sub> = 100
\n100gm of CaCO3<\/sub> = 1 Mole of CaCO3 <\/sub><\/p>\n

<\/a><\/div>\n
<\/a><\/div>\n<\/div>\n

Q. Calculate the number of atoms of Ca, C and O present in 150gm of CaCO3<\/sub>.<\/b><\/i>
\nSolution:
\nMolecular weight of CaCO3<\/sub> = 100
\nWeight of\u00a0 CaCO3<\/sub> = 150gram
\nNumber of Moles = ?
\nNumber of Molecules = ?
\nNumber of atoms of Ca, C and O = ?<\/p>\n

<\/a><\/div>\n

Number of molecules = Number of moles \u00d7 6.023\u00d71023<\/sup>
\n= 1.5\u00d76.023\u00d71023<\/sup>
\n= 9.0345\u00d71023<\/sup>
\n1 molecule of\u00a0 CaCO3<\/sub> contains 1 atom of Ca.
\n9.0345\u00d71023<\/sup> molecule of CaCO3<\/sub> contains 9.0345\u00d71023<\/sup> atoms of Ca.
\n1 molecule of CaCO3<\/sub> contains 1 atom of C.
\n9.0345\u00d71023<\/sup> molecule of CaCO3<\/sub> contains 9.0345\u00d71023<\/sup> atoms of C.
\n1 molecule of CaCO3<\/sub> contains 3 atoms of Oxyen.
\n9.0345\u00d71023<\/sup> molecule of CaCO3<\/sub> contains 3\u00d79.0345\u00d71023<\/sup> atoms of C.
\n= 2.7103\u00d71023<\/sup> atoms of C.<\/p>\n

Q. Calculate the number of moles and number of atoms of oxygen present in 80gm of oxygen.<\/b><\/i>
\nSolution:
\nMolecular Weight of Oxygen = 32
\nWeight of Oxygen = 80gm<\/p>\n

<\/a><\/div>\n

\u00a0<\/b><\/i><\/p>\n

Number of molecules = 2.5 \u00d7 6.023\u00d71023
\n= 1.5058\u00d71024<\/p>\n

One molecule of oxygen contains 2 atoms of oxyegn.
\n1.5058\u00d71024<\/sup> molecules of oxygen contains 2\u00d71.5058\u00d71024<\/sup>
\n= 3.01\u00d71024<\/sup> atoms of oxygen<\/p>\n

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

Molecular Weight is the number which shows how many times a molecule of a substance is heavier than 1\/12th part of an atom of C12 isotope. In other words molecular weight is the sum of atomic weight of all atoms present in a molecule of a substance. eg. Mol. wt. of CaCO3 =1\u00d740 + 1\u00d712 […]<\/p>\n","protected":false},"author":1,"featured_media":706,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_uf_show_specific_survey":0,"_uf_disable_surveys":false,"footnotes":""},"categories":[110],"tags":[],"class_list":["post-109","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-chemistry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/rijankc.com.np\/index.php?rest_route=\/wp\/v2\/posts\/109","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/rijankc.com.np\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/rijankc.com.np\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/rijankc.com.np\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/rijankc.com.np\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=109"}],"version-history":[{"count":1,"href":"https:\/\/rijankc.com.np\/index.php?rest_route=\/wp\/v2\/posts\/109\/revisions"}],"predecessor-version":[{"id":672,"href":"https:\/\/rijankc.com.np\/index.php?rest_route=\/wp\/v2\/posts\/109\/revisions\/672"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/rijankc.com.np\/index.php?rest_route=\/wp\/v2\/media\/706"}],"wp:attachment":[{"href":"https:\/\/rijankc.com.np\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=109"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/rijankc.com.np\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=109"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/rijankc.com.np\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=109"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}