The density of 2-methyl-2-butanol is 0.806 g\/mL. How many mgs represent an aliquot of 430 ML? How many mmols represent an aliquot of 0.650 mL (2-methyl-2-butanol, formula weight is 88.15 g\/mol)?<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The density of 2-methyl-2-butanol is 0.806 g\/mL. How many mgs represent an aliquot of 430 ML? How many mmols represent an aliquot of 0.650 mL (2-methyl-2-butanol, formula weight is 88.15 g\/mol)? The Correct Answer and Explanation is : Answer: Explanation:<\/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-184757","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/184757","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=184757"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/184757\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=184757"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=184757"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=184757"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Answer:<\/h3>\n\n\n\n
\n
\\text{Density} = \\frac{\\text{Mass}}{\\text{Volume}} \\quad \\text{so, Mass} = \\text{Density} \\times \\text{Volume}
]
[
\\text{Mass} = 0.806 \\, \\text{g\/mL} \\times 430 \\, \\text{mL} = 346.58 \\, \\text{g}
]
Converting to milligrams:
[
346.58 \\, \\text{g} \\times 1000 \\, \\text{mg\/g} = 346,580 \\, \\text{mg}
]
Answer: 346,580 mg<\/strong><\/li>\n\n\n\n
[
\\text{Mass} = 0.806 \\, \\text{g\/mL} \\times 0.650 \\, \\text{mL} = 0.5239 \\, \\text{g}
]
Convert to moles using the formula weight ((88.15 \\, \\text{g\/mol})):
[
\\text{Moles} = \\frac{\\text{Mass}}{\\text{Formula weight}} = \\frac{0.5239 \\, \\text{g}}{88.15 \\, \\text{g\/mol}} = 0.00594 \\, \\text{mol}
]
Converting to millimoles:
[
0.00594 \\, \\text{mol} \\times 1000 \\, \\text{mmol\/mol} = 5.94 \\, \\text{mmol}
]
Answer: 5.94 mmol<\/strong><\/li>\n<\/ol>\n\n\n\n
\n\n\n\nExplanation:<\/h3>\n\n\n\n
\n
Density connects the mass of a substance to its volume. By multiplying the density (( \\text{g\/mL} )) by the volume (( \\text{mL} )), the mass (( \\text{g} )) can be calculated. This is crucial for determining how much substance (by weight) is in a given volume.<\/li>\n\n\n\n
Since the question asks for milligrams, grams are converted to milligrams by multiplying by 1000 ((1 \\, \\text{g} = 1000 \\, \\text{mg})).<\/li>\n\n\n\n
To calculate the number of moles of a substance, the mass is divided by its molecular weight (formula weight, ( \\text{g\/mol} )). For smaller quantities, millimoles (( \\text{mmol} )) are used, which is equivalent to multiplying the moles by 1000.<\/li>\n\n\n\n
Such calculations are commonly used in chemistry and biochemistry for preparing solutions and understanding reactant quantities in reactions, ensuring precise experimental results.<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"