Question

What is the freezing point of a solution that contains 12.0 g of glucose (C6H12O6) in 50 g of acetic acid (CH3COOH). For acetic acid, Kf is 3.90°C/m and the melting point is 16.6 °C

Answer 1

Answer:

T = 11.4°C

Explanation:

Given that:

weight of solute  w (glucose) = 12 g

weight of solvent W (acetic acid) = 50 g

Molar depression constant of solvent (Kf)= 3.90°C/m

Temperature of solvent freezing point T₀ = 16.6°C

Temperature of solution freezing point Tₓ = (T₀ -T) = ???

Molar mass of solute (glucose) = 180.2 g/mol

Using the expression;

(T₀ -T) = $\frac{1000*k_f*w}{MM*W}$

$(16.6-T)=\frac{1000*3.90*12}{180.2*50}$

$(16.6-T)=\frac{46800}{9010}$

$(16.6-T) = 5.194^0C$

$(16.6-T) = 5.2^0C$

-T = -16.6 °C + 5.2 °C

- T = -11.4°C

T = 11.4°C

∴ The freezing point of the solution is said to be = 11.4°C