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3 years ago

Boiling point of a solution formed when 15.2 grams of CaCl2 dissolves in 57.0 g of water. kB= 0.512 c/m.

1 answer:

Nookie1986 [14]3 years ago

4 0

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 reaction

CaCl2 + 2H20 ⇒ Ca(OH)2 + HCl

given:

mass of CaCl2 = 15.2 grams

mass of the solution = 57 grams

Kb (molal elevation constant) = 0.512 c/m

i = vont hoff factor is 1 as 1 mole of the substance is given as product.

Molality is calculated as:

molality = $\frac{grams of solute}{grams of solution}$

= $\frac{15.2}{57}$

= 0.26 M

Boiling 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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Answer:

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3 years ago

The [OH-] of an aqueous solution is 3.5 x 10-5. What is the [H+]?

Sergeeva-Olga [200]

Answer: The concentration of hydrogen ion in the given solution is $2.81 \times 10^{-10}$ .

Explanation:

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$pOH = -log [OH^{-}]\\= -log (3.5 \times 10^{-5})\\= 4.45$

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pH + pOH = 14

pH + 4.45 = 14

pH = 14 - 4.45

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And, pH is the negative logarithm of concentration of hydrogen ions. Hence, $[H^{+}]$ is calculated as follows.

$pH = -log [H^{+}]\\9.55 = -log [H^{+}]\\H^{+} = 2.81 \times 10^{-10}$

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2 years ago

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