{"id":198552,"date":"2025-03-10T06:49:25","date_gmt":"2025-03-10T06:49:25","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=198552"},"modified":"2025-03-10T06:49:27","modified_gmt":"2025-03-10T06:49:27","slug":"200-grams-of-mgcl2-are-added-to-1-l-of-liquid-water-at-standard-conditions","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/10\/200-grams-of-mgcl2-are-added-to-1-l-of-liquid-water-at-standard-conditions\/","title":{"rendered":"200 grams of MgCl2 are added to 1 L of liquid water at standard conditions"},"content":{"rendered":"\n

200 grams of MgCl2 are added to 1 L of liquid water at standard conditions. a) Calculate the temperature at which water will freeze at 1 atm. b) Calculate the temperature at which water will boil at 1 atm
200 grams of MgCl2 are added to 1 L of liquid water at standard conditions.
a) Calculate the temperature at which water will freeze at 1 atm.
b) Calculate the temperature at which water will boil at 1 atm<\/p>\n\n\n\n

The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n

To solve this problem, we will use the freezing point depression<\/strong> and boiling point elevation<\/strong> formulas.<\/p>\n\n\n\n

Step 1: Determine the number of moles of MgCl\u2082<\/strong><\/h3>\n\n\n\n

The molar mass of MgCl\u2082<\/strong>:
[
\\text{Mg} = 24.305 \\text{ g\/mol}, \\quad \\text{Cl} = 2 \\times 35.453 \\text{ g\/mol}
]
[
= 24.305 + 70.906 = 95.211 \\text{ g\/mol}
]<\/p>\n\n\n\n

Moles of MgCl\u2082<\/strong> in 200 g:
[
\\frac{200 \\text{ g}}{95.211 \\text{ g\/mol}} = 2.1 \\text{ moles}
]<\/p>\n\n\n\n

Since MgCl\u2082<\/strong> dissociates into Mg\u00b2\u207a<\/strong> and 2 Cl\u207b<\/strong>, the number of particles produced:
[
\\text{van’t Hoff factor}, i = 3
]
Total moles of particles:
[
2.1 \\times 3 = 6.3 \\text{ moles of particles}
]<\/p>\n\n\n\n

Step 2: Freezing Point Depression<\/strong><\/h3>\n\n\n\n

The formula is:
[
\\Delta T_f = i K_f m
]
where:<\/p>\n\n\n\n

    \n
  • ( K_f )<\/strong> = 1.86 \u00b0C\u00b7kg\/mol (freezing point depression constant for water)<\/li>\n\n\n\n
  • ( m )<\/strong> = molality = (moles of particles) \/ (kg of solvent)<\/strong><\/li>\n<\/ul>\n\n\n\n

    Water mass = 1 kg<\/strong> (since 1 L \u2248 1 kg for water)<\/p>\n\n\n\n

    [
    m = \\frac{6.3}{1} = 6.3 \\text{ m}
    ]<\/p>\n\n\n\n

    [
    \\Delta T_f = 3 \\times 1.86 \\times 6.3 = 35.1^\\circ \\text{C}
    ]<\/p>\n\n\n\n

    New freezing point:
    [
    T_f = 0^\\circ C – 35.1^\\circ C = -35.1^\\circ C
    ]<\/p>\n\n\n\n

    Step 3: Boiling Point Elevation<\/strong><\/h3>\n\n\n\n

    The formula is:
    [
    \\Delta T_b = i K_b m
    ]
    where:<\/p>\n\n\n\n

      \n
    • ( K_b )<\/strong> = 0.512 \u00b0C\u00b7kg\/mol (boiling point elevation constant for water)<\/li>\n<\/ul>\n\n\n\n

      [
      \\Delta T_b = 3 \\times 0.512 \\times 6.3 = 9.67^\\circ \\text{C}
      ]<\/p>\n\n\n\n

      New boiling point:
      [
      T_b = 100^\\circ C + 9.67^\\circ C = 109.67^\\circ C
      ]<\/p>\n\n\n\n

      Final Answer:<\/strong><\/h3>\n\n\n\n

      a) Freezing Point<\/strong> = -35.1\u00b0C<\/strong>
      b) Boiling Point<\/strong> = 109.67\u00b0C<\/strong><\/p>\n\n\n\n

      \"\"<\/figure>\n\n\n\n

      The image visually represents the change in freezing and boiling points due to the addition of MgCl\u2082. You can see that the freezing point drops significantly to -35.1\u00b0C, while the boiling point rises to 109.67\u00b0C. <\/p>\n","protected":false},"excerpt":{"rendered":"

      200 grams of MgCl2 are added to 1 L of liquid water at standard conditions. a) Calculate the temperature at which water will freeze at 1 atm. b) Calculate the temperature at which water will boil at 1 atm200 grams of MgCl2 are added to 1 L of liquid water at standard conditions.a) Calculate the […]<\/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-198552","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/198552","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=198552"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/198552\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=198552"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=198552"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=198552"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}