The depression in freezing point observed for a formic acid solution of concentration 0.5 mL L⁻¹ is 0.0405°C. Density of formic acid is 1.05 g mL⁻¹. The Van't Hoff factor of the formic acid solution is nearly:

(Given for water $k_f = 1.86\ \mathrm{K\ kg\ mol^{-1}}$ )

Question

The depression in freezing point observed for a formic acid solution of concentration 0.5 mL L⁻¹ is 0.0405°C. Density of formic acid is 1.05 g mL⁻¹. The Van't Hoff factor of the formic acid solution is nearly:

(Given for water $k_f = 1.86\ \mathrm{K\ kg\ mol^{-1}}$ )

Paaras Thakur · Accepted Answer

1.9. Strategy: Calculate the mass of formic acid in the solution, find its molality, and then use the freezing point depression formula to find the van't Hoff factor.Step 1: Calculate mass and moles of formic acid (HCOOH)- Volume = 0.5 mL. Density = 1.05 g/mL.- Mass = 0.5 \ 1.05 = 0.525\ g.- Molar mass of HCOOH = 1 + 12 + 2(16) + 1 = 46\ g/mol.\{Moles } (n) = \{0.525}{46} 0.0114\{ mol}Step 2: Find Molality (m) 1 L of water (density 1 g/mL) is nearly 1 kg. 0.0114\{ mol/kg}Step 3: Determine i Tf = 0.0405\ K, Kf = 1.86\ K kg/mol. T_f = i K_f m 0.0405 = i \ 1.86 \ 0.0114 = \{0.0405}{0.0212} 1.91{\{Answer: (3)}}

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0.8

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1.1

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1.9

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2.4

The depression in freezing point observed for a formic acid solution of concentration 0.5 mL L⁻¹ is 0.0405°C. Density…

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