Answer :
Answer:
C C2H5OH = 7.598 molal
Explanation:
- molality (m) C2H5OH = n C2H5OH / Kg H2O
∴ V C2H5OH = 50.0 mL
∴ V H2O = 112.7 mL
∴ T = 20 °C
∴ δ C2H5OH (20°C) = 0.789 g/mL
∴ δ H2O (20°C) = 1.00 g/mL
∴ Mw C2H5OH = 46.0684 g/mol
⇒ n C2H5OH = (50.0 mL)×(0.789 g/mL)×(mol/46.0684 g) = 0.8563 mol
⇒ mass H2O = (112.7 mL)×(1.00 g/mL)×(Kg / 1000 g) = 0.1127 Kg
⇒ C C2H5OH = 0.8563 mol C2H5OH / 0.1127 Kg H2O
⇒ C C2H5OH = 7.598 molal
Molality is the measure of number of moles of solute particles in 1 kg of solution. Molality of the the given solution is 7.958 m.
Given here,
mass of Ethyl alcohol = 50 mL =( 50 x 0.789 ) = 39. 45 g
mass of Water = 112.7 mL = (112.7 x 1.00) = 112.7 gm
Molallity can be calculated by the formula,
[tex]\bold{Molality = \dfrac {w \times 1000}{m\times v}}[/tex]
Where,
w - mass of solute = 39. 45 g
m - molar mass of solute = 46.07
v - volume of solvent = 112.7 gm
put the values in the formula,
[tex]\bold{Molality = \dfrac {39.45 \times 1000}{46.07 \times 112.7}}\\\\\bold{Molality = 7.958}[/tex]
Therefore, the molality of the the given solution is 7.958 m.
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