Answer:
ethylene glycol (molar mass = 62.07 g/mol).
Explanation:
We can solve this problem using the relation:
ΔTf = (Kf)(m),
where, ΔTf is the depression in the freezing point (ΔTf = 25.5°C - 15.3°C = 10.2 °C).
Kf is the molal freezing point depression constant of water = 9.1 °C/m,
m is the molality of the solution (m = ??? m).
∴ m = ΔTf/(Kf) = (10.2 °C)/(9.1 °C/m) = 1.12 m.
Molality (m) is the no. of moles of solute per 1.0 kg of the solute.
m = (mass/molar mass) of the solute x (1000/mass of the solvent).
∴ 1.12 m = (0.807 g / molar mass of the solute) x (1000 / 11.6 g)
∴ molar mass of the solute = (0.807 g)(1000)/(1.12 m)(11.6 g) = 62.11 g/mol.
so, the unknown liquid is ethylene glycol (molar mass = 62.07 g/mol).
Tert-Butyl alcohol is the simplest form of tertiary alcohol, with the formula [tex]\bold{(CH_3)_3COH}[/tex]colorless non-electrolyte liquid that was dissolved in 11.6 g of tert-butyl alcohol is ethylene glycol (molar mass = 62.07 g/mol).
The correct option is A.
Tert-Butyl alcohol is colorless that melts at near room temperature. It is one of the four isomers of butanol.
Given,
The molar mass of ethylene glycol is 62.07 g/mol
KF is [tex]9.10^\circ C/m[/tex]
Mass of unknown liquid is 0.807 g
By the formula of the freezing point of depression
ΔTf = (Kf)(m)
ΔTf = 25.5 °C - 15.3 °C = 10.2 °C
Calculating the molality of the:
[tex]\bold{m = \dfrac{10.2^\circ C}{9.1^\circ C/m} = 1.12 m.}[/tex]
Molality (m) is the no. of moles of solute per 1.0 kg of the solute.
[tex]\bold{m =\dfrac{mass}{molar\; mass} \times \dfrac{1000}{mass\; of\; the \;solvent} }\\\\\\\bold{m =\dfrac{0.807 g}{1000} \times \dfrac{1.12 m}{11.6 g} =62.11 g/mol}[/tex]
Thus, the unknown non-electrolyte bis ethylene glycol (molar mass = 62.07 g/mol).
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