{"id":29419,"date":"2025-06-21T03:53:46","date_gmt":"2025-06-21T03:53:46","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=29419"},"modified":"2025-06-21T03:53:48","modified_gmt":"2025-06-21T03:53:48","slug":"the-molality-of-a-solution-denoted-here-by-the-symbol-m-is-a-measure-of-concentration-defined-as-moles-of-solute-divided-by-mass-of-solvent-or-in-other-words-m-n_molute-mn_solvent-where-m-is","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-molality-of-a-solution-denoted-here-by-the-symbol-m-is-a-measure-of-concentration-defined-as-moles-of-solute-divided-by-mass-of-solvent-or-in-other-words-m-n_molute-mn_solvent-where-m-is\/","title":{"rendered":"The molality of a solution (denoted here by the symbol m) is a measure of concentration, defined as moles of solute divided by mass of solvent, or in other words m = (n_molute)\/(Mn_solvent) where M is the solvent’s molar mass."},"content":{"rendered":"\n

The molality of a solution (denoted here by the symbol m) is a measure of concentration, defined as moles of solute divided by mass of solvent, or in other words m = (n_molute)\/(Mn_solvent) where M is the solvent’s molar mass. For aqueous solutions of NaCl at 25.0\u00b0C and 1 atm, and for 0 <= m <= 6 mol\/(kg), the available volumetric data can be fit to the empirical function V = c0 + c1m + c2m^2 + c3m^3 with parameters c0 = 1001.70 cm^3 c1 = 17.298 kgcm^3\/mol c2 = 0.9777 kg^2cm^3\/mol^2 c3 = -0.0569 kg^3cm^3\/mol^3. (a) Use this expression for V to derive an expression for the partial molar volume of NaCl. (b) Determine the partial molar volume of water (again in aqueous NaCl) as a function of m. (c) Given that the density of pure water (at 25\u00b0C and 1 atm) is 0.99707 g\/cm^3, calculate the partial molar volume of water in a 1.250 molal solution of NaCl. What is the fractional change in hat(V)_water in this solution, relative to hat(V)_water? (d) Given 1.00 kg of a 1.250 molal solution of aqueous NaCl, suppose that an additional 10.0 g of NaCl is dissolved into this solution. Calculate the change in the volume of the solution. (e) What is the molar concentration (moles of NaCl per liter of solution) of a solution that is 1.250 molal sodium chloride?<\/p>\n\n\n\n

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

Let’s address each part step by step, clearly and concisely:<\/p>\n\n\n\n

\n\n\n\n

(a) Expression for the Partial Molar Volume of NaCl<\/strong><\/h3>\n\n\n\n

We are given: V(m)=c0+c1m+c2m2+c3m3V(m) = c_0 + c_1m + c_2m^2 + c_3m^3V(m)=c0\u200b+c1\u200bm+c2\u200bm2+c3\u200bm3<\/p>\n\n\n\n

where VVV is the volume of the solution per kg of solvent (in cm\u00b3), and mmm is the molality in mol\/kg.<\/p>\n\n\n\n

The partial molar volume of NaCl<\/strong>, V^NaCl\\hat{V}_{\\text{NaCl}}V^NaCl\u200b, is given by: V^NaCl=(\u2202V\u2202m)T,P\\hat{V}_{\\text{NaCl}} = \\left( \\frac{\\partial V}{\\partial m} \\right)_{T,P}V^NaCl\u200b=(\u2202m\u2202V\u200b)T,P\u200b<\/p>\n\n\n\n

Differentiate VVV with respect to mmm: V^NaCl=dVdm=c1+2c2m+3c3m2\\hat{V}_{\\text{NaCl}} = \\frac{dV}{dm} = c_1 + 2c_2m + 3c_3m^2V^NaCl\u200b=dmdV\u200b=c1\u200b+2c2\u200bm+3c3\u200bm2<\/p>\n\n\n\n

\n\n\n\n

(b) Partial Molar Volume of Water<\/strong><\/h3>\n\n\n\n

Let VVV be total volume per 1 kg solvent. The total volume can be written as: V=nNaClV^NaCl+nH2OV^H2OV = n_{\\text{NaCl}} \\hat{V}_{\\text{NaCl}} + n_{\\text{H}_2O} \\hat{V}_{\\text{H}_2O}V=nNaCl\u200bV^NaCl\u200b+nH2\u200bO\u200bV^H2\u200bO\u200b<\/p>\n\n\n\n

Given nNaCl=mn_{\\text{NaCl}} = mnNaCl\u200b=m and nH2O=1000\/Mn_{\\text{H}_2O} = 1000\/MnH2\u200bO\u200b=1000\/M where M=18.015 g\/molM = 18.015 \\text{ g\/mol}M=18.015 g\/mol, then: V^H2O=V\u2212mV^NaClnH2O=V\u2212mV^NaCl1000\/18.015\\hat{V}_{\\text{H}_2O} = \\frac{V – m\\hat{V}_{\\text{NaCl}}}{n_{\\text{H}_2O}} = \\frac{V – m\\hat{V}_{\\text{NaCl}}}{1000\/18.015}V^H2\u200bO\u200b=nH2\u200bO\u200bV\u2212mV^NaCl\u200b\u200b=1000\/18.015V\u2212mV^NaCl\u200b\u200b<\/p>\n\n\n\n

\n\n\n\n

(c) Partial Molar Volume of Water at 1.250 molal<\/strong><\/h3>\n\n\n\n

First calculate VVV: V=1001.70+17.298(1.250)+0.9777(1.250)2\u22120.0569(1.250)3=1024.46 cm3V = 1001.70 + 17.298(1.250) + 0.9777(1.250)^2 – 0.0569(1.250)^3 = 1024.46 \\text{ cm}^3V=1001.70+17.298(1.250)+0.9777(1.250)2\u22120.0569(1.250)3=1024.46 cm3<\/p>\n\n\n\n

Now use: V^NaCl=c1+2c2m+3c3m2=17.298+2(0.9777)(1.25)+3(\u22120.0569)(1.25)2=19.56 cm3\/mol\\hat{V}_{\\text{NaCl}} = c_1 + 2c_2m + 3c_3m^2 = 17.298 + 2(0.9777)(1.25) + 3(-0.0569)(1.25)^2 = 19.56 \\text{ cm}^3\/molV^NaCl\u200b=c1\u200b+2c2\u200bm+3c3\u200bm2=17.298+2(0.9777)(1.25)+3(\u22120.0569)(1.25)2=19.56 cm3\/mol V^H2O=1024.46\u22121.250(19.56)1000\/18.015=1024.46\u221224.4555.51\u224818.01 cm3\/mol\\hat{V}_{\\text{H}_2O} = \\frac{1024.46 – 1.250(19.56)}{1000\/18.015} = \\frac{1024.46 – 24.45}{55.51} \\approx 18.01 \\text{ cm}^3\/molV^H2\u200bO\u200b=1000\/18.0151024.46\u22121.250(19.56)\u200b=55.511024.46\u221224.45\u200b\u224818.01 cm3\/mol<\/p>\n\n\n\n

For pure water: V=1000 g0.99707 g\/cm3=1002.94 cm3,V^H2O=1002.9455.51\u224818.07 cm3\/molV = \\frac{1000 \\text{ g}}{0.99707 \\text{ g\/cm}^3} = 1002.94 \\text{ cm}^3, \\quad \\hat{V}_{\\text{H}_2O} = \\frac{1002.94}{55.51} \\approx 18.07 \\text{ cm}^3\/molV=0.99707 g\/cm31000 g\u200b=1002.94 cm3,V^H2\u200bO\u200b=55.511002.94\u200b\u224818.07 cm3\/mol<\/p>\n\n\n\n

Fractional change<\/strong>: 18.01\u221218.0718.07\u2248\u22120.0033 or \u22120.33%\\frac{18.01 – 18.07}{18.07} \\approx -0.0033 \\text{ or } -0.33\\%18.0718.01\u221218.07\u200b\u2248\u22120.0033 or \u22120.33%<\/p>\n\n\n\n

\n\n\n\n

(d) Volume Change After Adding 10.0 g NaCl to 1.250 molal Solution<\/strong><\/h3>\n\n\n\n

Original moles NaCl = 1.250 mol
Add 10.0 g NaCl \u2192 10.058.44\u22480.1711\\frac{10.0}{58.44} \\approx 0.171158.4410.0\u200b\u22480.1711 mol
New total = 1.4211 mol
New molality = 1.4211 mol\/kg<\/p>\n\n\n\n

Use: \u0394V=V(1.4211)\u2212V(1.250)\\Delta V = V(1.4211) – V(1.250)\u0394V=V(1.4211)\u2212V(1.250)<\/p>\n\n\n\n

Compute: V(1.4211)=1001.70+17.298(1.4211)+0.9777(1.4211)2\u22120.0569(1.4211)3\u22481027.23 cm3V(1.4211) = 1001.70 + 17.298(1.4211) + 0.9777(1.4211)^2 – 0.0569(1.4211)^3 \\approx 1027.23 \\text{ cm}^3V(1.4211)=1001.70+17.298(1.4211)+0.9777(1.4211)2\u22120.0569(1.4211)3\u22481027.23 cm3 \u0394V=1027.23\u22121024.46=2.77 cm3\\Delta V = 1027.23 – 1024.46 = 2.77 \\text{ cm}^3\u0394V=1027.23\u22121024.46=2.77 cm3<\/p>\n\n\n\n

\n\n\n\n

(e) Molar Concentration of 1.250 molal NaCl<\/strong><\/h3>\n\n\n\n

We know:<\/p>\n\n\n\n

    \n
  • 1.250 mol NaCl<\/li>\n\n\n\n
  • Volume from (c): 1024.46 cm\u00b3 = 1.02446 L<\/li>\n<\/ul>\n\n\n\n

    So: Molarity=1.250 mol1.02446 L\u22481.22 mol\/L\\text{Molarity} = \\frac{1.250 \\text{ mol}}{1.02446 \\text{ L}} \\approx 1.22 \\text{ mol\/L}Molarity=1.02446 L1.250 mol\u200b\u22481.22 mol\/L<\/p>\n\n\n\n

    \n\n\n\n

    Summary<\/strong>:<\/p>\n\n\n\n

      \n
    • V^NaCl=c1+2c2m+3c3m2\\hat{V}_{\\text{NaCl}} = c_1 + 2c_2m + 3c_3m^2V^NaCl\u200b=c1\u200b+2c2\u200bm+3c3\u200bm2<\/li>\n\n\n\n
    • V^H2O=V\u2212mV^NaClnH2O\\hat{V}_{\\text{H}_2O} = \\frac{V – m\\hat{V}_{\\text{NaCl}}}{n_{\\text{H}_2O}}V^H2\u200bO\u200b=nH2\u200bO\u200bV\u2212mV^NaCl\u200b\u200b<\/li>\n\n\n\n
    • V^H2O\\hat{V}_{\\text{H}_2O}V^H2\u200bO\u200b decreases slightly (\u22120.33%) in 1.250 molal solution<\/li>\n\n\n\n
    • Adding 10 g NaCl increases volume by ~2.77 cm\u00b3<\/li>\n\n\n\n
    • Molarity of 1.250 molal NaCl \u2248 1.22 mol\/L<\/li>\n<\/ul>\n\n\n\n
      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      The molality of a solution (denoted here by the symbol m) is a measure of concentration, defined as moles of solute divided by mass of solvent, or in other words m = (n_molute)\/(Mn_solvent) where M is the solvent’s molar mass. For aqueous solutions of NaCl at 25.0\u00b0C and 1 atm, and for 0 <= m […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-29419","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29419","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=29419"}],"version-history":[{"count":2,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29419\/revisions"}],"predecessor-version":[{"id":29422,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29419\/revisions\/29422"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=29419"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=29419"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=29419"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}