An aqueous CaCl2 solution has a vapor pressure of 83.5 mmHg at 50 C. The vapor pressure of pure water at this temperature is 92.6 mmHg. What is the concentration of CaCl2 in mass percent? Please provide formula used and some details. Thanks.
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Answer:83.5 = 92.6 X X = moles fraction water =0.902 = moles water / moles water + moles CaCl2 1 - 0.902 = 0.098 = moles CaCl2 / moles CaCl2 + moles water
An aqueous CaCl2 solution has a vapor pressure of 84.0 mmHg at 50 degrees C. The vapor pressure of pure H2O at this temperature is 92.6 mmHg. What is the concentration of CaCl2 in mass percent?
Answer:p(total) = 176.6 mmHg x(water) = 84/176.6 = 0.48 x(cacl2) = 1-0.48 = 0.52 n(water) = n(total)*x(water) n(cacl2) = n(total)*x(cacl2) %mass(cacl2) = 100 * (mass(cacl2)/mass(total)) mass(cacl2)= n(cacl2)*M(cacl2) mass(water)=n(water)*M(water) mass(total)=m(water)+m(cacl2) n(water)/n(cacl2)= 0.92 m(w)/m(cacl2) = 0.92 * (18/111) =0.15 m(w) = 0.15*m(cacl2) m(total) = 0.15cacl2 + cacl2 = 1.15cacl2 (1/1.15)*100 = %86.96
An aqueous CaCl2 solution has a vapor pressure of 82.2 mmHg at 50 degrees C. The vapor pressure of pure water at this temperature is 92.6 mmHg. What is the concentration of CaCl2 in mass percent?
Answer:From Raoult's law:Vapor pressure of solution = mole fraction of water x vapor pressure of watermole fraction of water = 82.2/92.6 = 0.8877mole fraction of Ca2+ and Cl- ions = 1 - 0.8877 = 0.1123For every 0.8877 moles of water, there are 0.1123/3 = 0.03743 moles of CaCl2 since CaCl2 => Ca2+ + 2Cl- and each CaCl2 gives 3 solute ions.Since mass = moles x molar mass,mass% CaCl2 = mass of CaCl2/(mass of CaCl2 + mass of H2O) x 100= (0.03743 x 110.98)/(0.03743 x 110.98 + 0.8877 x 18.02) x 100= 20.6%
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