Boiling point of a solution formed when 15.2 grams of CaCl2 dissolves in 57.0 g of water. kB= 0.512 c/m.
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Question:
Boiling point of a solution formed when 15.2 grams of CaCl2 dissolves in 57.0 g of water. kB= 0.512 c/m.
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100.133 degree celsius is the boiling point of the solution formed when 15.2 grams of CaCl2 dissolves in 57.0 g of water. Explanation:Balanced eaquation for the reactionCaCl2 + 2H20 ⇒ Ca(OH)2 + HClgiven:mass of CaCl2 = 15.2 gramsmass of the solution = 57 gramsKb (molal elevation constant) = 0.512 c/mi = vont hoff factor is 1 as 1 mole of the substance is given as product.Molality is calculated as:molality = [tex]\frac{grams of solute}{grams of solution}[/tex] = [tex]\frac{15.2}{57}[/tex] = 0.26 MBoiling point is calculated as:ΔT = i x Kb x M = 1 x 0.512 x 0.26 = 0.133 degrees The boiling point of the solution will be:100 degrees + 0.133 degrees (100 degrees is the boiling point of water)= 100.133 degree celcius is the boiling point of mixture formed.
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