The normal freezing temperature for pure water is 0c. Howeverif sugar is added in the pure water, the freezing point will be lower than zero. How far below zero will depend on the sugar concentration in the water.
-5.58 C
The freezing point depression constant for water is approximately 1.858°C kg/mol. Using the formula ∆Tf = iKfm, where i is the van't Hoff factor, Kf is the freezing point depression constant, and m is the molality, we can calculate the change in freezing point. In this case, with 4 mol of sugar added to 1 kg of water (1000 g), the molality is 4 mol / 1000 g = 4 m/kg. Plugging these values in, we get ∆Tf = 1 * 1.858 * 4 = 7.432°C. The freezing point of water would decrease by 7.432°C.
7.44 C
-5.58 C
it takes about 6 hours
The freezing point of water decreases by about 1.86 degrees Celsius for each mole of solute (such as sugar) dissolved in 1 kg of water. So, the freezing point would decrease by 1.86 degrees Celsius for every mole of sugar added.
7.44 degrees c.
Adding 4 mol sugar to 1 g (gram) water is impossible !
7.44°C~apex
-5.58 C
Any impurity decrease the freezing point of water.
The normal freezing temperature for pure water is 0c. Howeverif sugar is added in the pure water, the freezing point will be lower than zero. How far below zero will depend on the sugar concentration in the water.
The freezing point depression constant for water is approximately 1.858°C kg/mol. Using the formula ∆Tf = iKfm, where i is the van't Hoff factor, Kf is the freezing point depression constant, and m is the molality, we can calculate the change in freezing point. In this case, with 4 mol of sugar added to 1 kg of water (1000 g), the molality is 4 mol / 1000 g = 4 m/kg. Plugging these values in, we get ∆Tf = 1 * 1.858 * 4 = 7.432°C. The freezing point of water would decrease by 7.432°C.
7.44 C
-5.58 C
-5.58 C
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