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3 years ago

What is the number of kilograms of solvent in a 0.70 molal solution containing 5.0 grams of solute?

1 answer:

3 0

Molar mass of solute is 30 g/mol
molality is defined as the number of moles of solute in 1 kg of solvent
number of moles of solute - 5.0 g / 30 g/mol = 0.17 mol
molality of solution is 0.70 mol/kg
the mass of solvent when there are 0.7 mol of solute - 1 kg
then the mass of solvent when there's 0.17 mol of solute - 0.17 mol / 0.7 mol/kg
= 0.242 kg
therefore mass of solvent is 0.242 kg

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3 years ago

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant o

sattari [20]

Answer:

4.26 %

Explanation:

There is some info missing. I think this is the original question.

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10 ⁻⁴.

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50 × 10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

$[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M$

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

$\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%$

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3 years ago

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kotykmax [81]

Answer:

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Explanation:

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