What is the freezing point of a solution of 7.15 g MgCl₂ in 100 g of water? K_f for water is $1.86 ^ { \circ } \mathrm { Cm } ^ { - 1 }$ A)-0.140 °C B)-1.40 °C C)-2.80 °C D)-4.19 °C

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The freezing point depression is calculated using the formula ΔTf = i * Kf * m, where i is the van't Hoff factor (3 for MgCl2), Kf is the cryoscopic constant (1.86 °C kg/mol for water), and m is the molality of the solution. First, calculate the molality: moles of MgCl2 = 7.15 g / 95.21 g/mol = 0.0751 mol; molality = 0.0751 mol / 0.1 kg = 0.751 m. Then, ΔTf = 3 * 1.86 °C kg/mol * 0.751 m = 4.19 °C. The freezing point is 0 °C - 4.19 °C = -4.19 °C.

What Is the Freezing Point of a Solution of 7 1.86∘Cm−11.86 ^ { \circ } \mathrm { Cm } ^ { - 1 }1.86∘Cm−1

What Is the Freezing Point of a Solution of 7 $1.86 ^ { \circ } \mathrm { Cm } ^ { - 1 }$