The concentration of sodium iodide used is 15% in acetone. Calculate the number of moles of sodium acetate used in each test tube. Assume the density of acetone = density of the solution. (Look up the density of acetone). Show your work. Density of Acetone= 0.788 g\/ml Molar mass of acetone= 58.08 2ml 15% NaI in acetone (0.788g \/ ml) * ( 1 mol \/ 58.08 g) = 0.027 moles<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s go step-by-step to calculate the number of moles of sodium iodide (NaI) used in each test tube.<\/p>\n\n\n\n Since the density of the solution is the same as acetone, we can calculate the mass of the solution in 2 mL. The mass (in grams) is given by: The concentration is 15% (w\/v), meaning there are 15 grams of NaI in every 100 mL of solution. To find the mass of NaI in 2 mL of solution, use the following relationship: To calculate the number of moles of NaI, we use the formula: The number of moles of sodium iodide used in each test tube (2 mL of 15% NaI in acetone) is 0.0020 moles<\/strong>.<\/p>\n\n\n\n In this problem, the first key step was calculating the mass of the solution in the test tube using the density and volume. Then, using the given concentration of NaI (15% w\/v), we determined the mass of NaI present in the 2 mL of solution. Finally, using the molar mass of NaI, we converted the mass of NaI into moles. This process allowed us to find the number of moles of sodium iodide in the 2 mL of acetone solution.<\/p>\n","protected":false},"excerpt":{"rendered":" The concentration of sodium iodide used is 15% in acetone. Calculate the number of moles of sodium acetate used in each test tube. Assume the density of acetone = density of the solution. (Look up the density of acetone). Show your work. Density of Acetone= 0.788 g\/ml Molar mass of acetone= 58.08 2ml 15% NaI […]<\/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-204189","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204189","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=204189"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204189\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=204189"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=204189"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=204189"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Given Data<\/h3>\n\n\n\n
\n
Step 2: Calculate the Mass of Solution in 2 mL<\/h3>\n\n\n\n
[
\\text{mass of solution} = \\text{density} \\times \\text{volume}
]
[
\\text{mass of solution} = 0.788 \\, \\text{g\/mL} \\times 2 \\, \\text{mL} = 1.576 \\, \\text{g}
]
So, the mass of the solution in 2 mL is 1.576 grams.<\/p>\n\n\n\nStep 3: Calculate the Mass of Sodium Iodide (NaI) in the Solution<\/h3>\n\n\n\n
[
\\text{mass of NaI} = \\frac{\\text{concentration of NaI}}{100} \\times \\text{volume of solution}
]
[
\\text{mass of NaI} = \\frac{15 \\, \\text{g}}{100 \\, \\text{mL}} \\times 2 \\, \\text{mL} = 0.3 \\, \\text{g}
]
Thus, the mass of NaI in 2 mL of solution is 0.3 grams.<\/p>\n\n\n\nStep 4: Calculate the Moles of Sodium Iodide (NaI)<\/h3>\n\n\n\n
[
\\text{moles of NaI} = \\frac{\\text{mass of NaI}}{\\text{molar mass of NaI}}
]
The molar mass of NaI is:
[
\\text{Molar mass of NaI} = 22.99 \\, \\text{g\/mol (for Na)} + 126.90 \\, \\text{g\/mol (for I)} = 149.89 \\, \\text{g\/mol}
]
Now, calculate the moles:
[
\\text{moles of NaI} = \\frac{0.3 \\, \\text{g}}{149.89 \\, \\text{g\/mol}} = 0.0020 \\, \\text{mol}
]<\/p>\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
Explanation:<\/h3>\n\n\n\n