Question 12 Marks
Calculate the temperature at which the solution containing $54\ g$ of glucose, $\ce{C_6H_{12}O_6}$ in $250\ g$ of water will freeze. $(K_b$ for water $= 1.86\ K \ kg\ mol^{-1})$
Answer
View full question & answer→Molecular mass of glucose
$M_B=72+12+96=180\ g\ mol^{-1}$
$\Delta T_f=\frac{K_f \times w_B \times 1000}{M_B \times w_A}$
$=\frac{1.86 \times 54 \times 1000}{180 \times 250}=2.23 K$
Freezing point of solution $= T _{ f }^0-\Delta T f=273-2.23=270.77 K$
$M_B=72+12+96=180\ g\ mol^{-1}$
$\Delta T_f=\frac{K_f \times w_B \times 1000}{M_B \times w_A}$
$=\frac{1.86 \times 54 \times 1000}{180 \times 250}=2.23 K$
Freezing point of solution $= T _{ f }^0-\Delta T f=273-2.23=270.77 K$
