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Solve the following.
Calculate the amount of CaCl₂ (molar mass = 111 g mol⁻¹) which must be added to 500 g of water to…
Calculate the amount of $\text{CaCl}_2$ (molar mass = $111\text{ g mol}^{-1}$) which must be added to $500\text{ g}$ of water to lower its freezing point by $2\text{ K}$, assuming $\text{CaCl}_2$ is completely dissociated. ($K_f\text{ for water} = 1\cdot86\text{ K kg mol}^{-1}$).
Answer
Answer
Formula used: Depression of freezing point: dTf = i*Kf*m with i = 3, Kf = 1.86 K kg mol^-1, M2 = 111 g mol^-1, w1 = 500 g, dTf = 2 K, so w2 = 19.89 g
$\mathrm{CaCl_2}$ dissociates into 3 ions, so $i = 3$.
$\Delta T_f = \dfrac{i K_f w_2 \times 1000}{M_2 w_1}$
$2 = \dfrac{3 \times 1\cdot86 \times w_2 \times 1000}{111 \times 500}$
$w_2 = \dfrac{2 \times 111 \times 500}{3 \times 1\cdot86 \times 1000} = 19\cdot89\ \mathrm{g}$
Final answer: 19.89 g