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stealth61 [152]
1 year ago
9

Calculate the molality of a 5.51 M ethanol (C2H5OH) solution whose density is 0.9349 g/mL

Chemistry
1 answer:
bearhunter [10]1 year ago
5 0

Molality of C2H5OH is 1.1.27m.



What is Molality?

Molality is no.of moles present in One Kg solution .

Molality is represented by m

m= no.of moles/ weigt of solution in kg

Given is Molarity= 51.30M

molar mass of C2H5OH is 46g / mol , Density =0.9349g/ ml

Density=m/V

V = 107 ml

Molarity= no of moles/ Volume

51.5×46×107= x × 1000

weight of C2H5OH = 27.7g

molality = 27.7/1000××46

m= 1.27m

to learn more about Molality click brainly.com/question/20380670

#SPJ9

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         P_{1} = 100 mm Hg or \frac{100}{760}atm = 0.13157 atm

         T_{1} = 1080 ^{o}C = (1080 + 273) K = 1357 K

         T_{2} = 1220 ^{o}C = (1220 + 273) K = 1493 K

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According to Clasius-Clapeyron equation,

                   log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}

            log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]

          log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]

                0.77815 = \frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K

              \Delta H_{vap} = 2.219 \times 10^{5} J/mol

                                   = 2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}

                                    = 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

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