Calculate the molality and van’t Hoff factor (i) for the following aqueous solutions:

1 answer:

3 0

The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

<h3>What is the value of Van t Hoff factor?</h3>

For most non-electrolytes dissolved in water, the Van 't Hoff factor is essentially $ 1 $ . For most ionic compounds dissolved in water, the Van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.

<h3>Which has highest Van t Hoff factor?</h3>

The Van't Hoff factor will be highest for

A. Sodium chloride.

B. Magnesium chloride.

C. Sodium phosphate.

D. Urea.

Learn more about van't off factor here:

<h3>brainly.com/question/22047232</h3><h3 /><h3>#SPJ4</h3>

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Suppose of ammonium nitrate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of ammonium c

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Answer:

Final molarity of ammonium cation in the solution = 0.16 M

Explanation:

Complete Question

Suppose 2.59 g of ammonium nitrate is dissolved in 200. mL of a 0.40M aqueous solution of sodium chromate. Calculate the final molarity of ammonium cation in the solution. You can assume the volume of the solution doesn't change when the ammonium nitrate is dissolved in it. Be sure your answer has the correct number of significant digits.

Solution

2NH₄NO₃ + Na₂CrO₄ → (NH₄)₂CrO₄ + 2NaNO₃

We first convert the given parameters to number of moles

Number of moles = (Mass/Molar mass)

Molar mass of NH₄NO₃ = 80.043 g/mol

Number of moles of NH₄NO₃ = (2.59/80.043) = 0.03224 mole

Number of moles = (Concentration in mol/L) × (Volume in L)

Number of moles of Na₂CrO₄ = 0.4 × 0.2 = 0.08 Mole

2 moles of NH₄NO₃ react with 1 mole of Na₂CrO₄

So, it it evident that NH₄NO₃ is the limiting reagent as it is in short supply in the amount needed for the reaction.

So, the number of moles of ammonium ion in the product is also 0.03224 mole.

Molarity = (Number of moles)/(Volume L)