Question

An aqueous solution has 5%urea and 10%glucose by weight. what will be the freezing point of this solution

Answer 1

Answer:The freezing point of this solution will be 269.97 K.

Explanation;

Suppose in ,100 gram of solution there are 5 g of urea, 10 g of glucose and 85 g.

Moles of urea = $n_1=\frac{5 g}{60.06 g/mol}=0.0832 moles$

Moles of glucose = $n_2=\frac{10 g}{180.156 g/mol}=0.0555 moles$

The molal depression constant of water = 1.86 K kg/mol

Mass of the water present in the solution = 85 g = 0.085 kg

$\Delta T=K_f\times \frac{n_1+n_2}{\text{mass pf solvent in kg}}$

$\Delta T=1.86 K kg/mol\times \frac{0.0832 moles+0.0555 moles}{0.085 kg}=3.03 K$

$\Delta T=T_{f}^o-T_f$

$3.03 k=273 k-T_f$

$T_f=269.97 K$

The freezing point of this solution will be 269.97 K.

Answer 2

Answer:

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