Given that the value of kf for water is 1.86°c/m, what is the freezing point of an aqueous solution of nacl prepared by mixing 427.0 g solute with 1.000 kg of water

Answer is: the freezing point of an aqueous solution of sodium chloride is -13.6°C.
m(NaCl) = 427.0 g.
n(NaCl) = m(NaCl) ÷ M(NaCl).
n(NaCl) = 427 g ÷ 58.4 g/mol.
n(NaCl) = 7.31 mol.
b(NaCl) = n(NaCl) ÷ m(H₂O).
b(NaCl) = 7.31 mol ÷ 1.00 kg.
b(NaCl) = 7.31 m.
ΔT = Kf · b(NaCl).
ΔT = 1.86°C/m · 7.31 m.
ΔT = 13.596°C.

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batolisis batolisis · Answer 2 · 2022-03-22T12:50:37+01:00

The freezing point of the aqueous solution of NaCl is : -27.2⁰c

Given data :

Kf = 1.86° c/m

mass of solute ( w ) = 427.0 g

mass of solvent ( water ) ( W ) = 1 kg = 1000 g

molecular mass of Nacl = 58.45 g/mol

Determine the freezing point

Applying the relation between depression in freezing point and molar depression constant

ΔTf = m * kf * i

also

ΔTf = ( kf * w * 1000) / ( W * m ) * i ------- ( 1 )

where : kf = 1.86° c/m, w = 427g, W = 1000 g, i = 2

Insert values into equation ( 1 )

Therefore:

ΔTf = 13.58 * 2 = 27.2 °c

Note : ΔTf = T - Ts

Ts ( freezing point ) = T - ΔTf

= 0 - 27.2 = -27.2 °c

Hence we can conclude that The freezing point of the aqueous solution of NaCl is : -27.2⁰c

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