How to Calculate Freezing Point Depression

The most effective method to Calculate Freezing Point Depression This model issue shows how to figure the point of solidification misery utilizing arrangement of salt in water. Speedy Review of Freezing Point Depression The point of solidification sorrow is one of the colligative properties of issue, which implies it is influenced by the quantity of particles, not the concoction personality of the particles or their mass. At the point when a solute is added to a dissolvable, its the point of solidification is brought down from the first estimation of the unadulterated dissolvable. It doesnt matter whether the solute is a fluid, gas, or strong. For instance, the point of solidification wretchedness happens when either salt or liquor are added to water. Indeed, the dissolvable can be any stage, as well. The point of solidification melancholy additionally happens in strong blends. The point of solidification misery is determined utilizing Raoults Law and the Clausius-Clapeyron Equation to compose a condition called Blagdens Law. In a perfect arrangement, the point of solidification despondency just relies upon solute fixation. The point of solidification Depression Problem 31.65 g of sodium chloride is added to 220.0 mL of water at 34  °C. By what means will this influence theâ freezing purpose of the water?Assume theâ sodium chloride totally separates in the water.Given: thickness of water at 35  °C 0.994 g/mLKf water 1.86  °C kg/molSolution:To find theâ temperature change rise of a dissolvable by a solute, utilize the point of solidification discouragement equation:ÃŽT iKfmwhereÃŽT Change in temperature in  °Ci van t Hoff factorKf molal the point of solidification wretchedness steady or cryoscopic consistent in  °C kg/molm molality of the solute in mol solute/kg solvent.Step 1 Calculate the molality of the NaClmolality (m) of NaCl moles of NaCl/kg waterFrom the occasional table, locate the nuclear masses of the elements:atomic mass Na 22.99atomic mass Cl 35.45moles of NaCl 31.65 g x 1 mol/(22.99 35.45)moles of NaCl 31.65 g x 1 mol/58.44 gmoles of NaCl 0.542 molkg water thickness x volumekg water 0.994 g/mL x 220 mL x 1 kg/1000 gkg water 0.219 kgmNaCl moles of NaCl/kg watermNaCl 0.542 mol/0.219 kgmNaCl 2.477 mol/kgStep 2 Determine the van t Hoff factorThe van t Hoff factor, I, is a consistent related with the measure of separation of the solute in the dissolvable. For substances which don't separate in water, for example, sugar, I 1. For solutes that totally separate intoâ two particles, I 2. For this model, NaCl totally separates into the two particles, Na and Cl-. Along these lines, I 2 for this example.Step 3 Find ÃŽTÃŽT iKfmÃŽT 2 x 1.86  °C kg/mol x 2.477 mol/kgÃŽT 9.21  °CAnswer:Adding 31.65 g of NaCl to 220.0 mL of water will bring down the point of solidification by 9.21  °C.