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monitta
2 years ago
11

What mass of solid that has a molar mass 89.0 g/mol should be added to 100.0 g of benzene to raise the boiling point of benzene

by 2.42°C? (The boiling point elevation constant of benzene is 2.53°C•kg/mol.) 2.84 g 8.52 g 21.5 g 93.0 g
Chemistry
1 answer:
taurus [48]2 years ago
3 0

On the basis of given data:

Mass of solvent (benzene) = 100 gm or 0.1 Kg

Molar mass of solute (solid) = 89 g/mol

Increase in boiling point (deltaTb) = 2.42 degree C

The boiling point elevation constant k = 2.53 C.kg/mol

There is a need to determine the mass of the solid to be added, the elevation in boiling point is proportional to the m (molality) of the solute;

ΔTb = k.m

Here, m that is molality = moles of solute/kg of solvent

Therefore,

ΔTb = k (benzene) × moles of solid/kg of benzene

2.42 = 2.53 × moles of solid/0.1

moles of solid = 0.0956 moles

The molar mass of solid = 89 g/mol

Thus, the mass of solid = 0.0956 moles × 89 g/mol

= 8.508 g

Thus, the mass of solid to be added is 8.508 g.

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lilavasa [31]

1.66 M is the concentration of the chemist's working solution.

<h3>What is molarity?</h3>

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.

In this case, we have a solution of Zn(NO₃)₂.

The chemist wants to prepare a dilute solution of this reactant.

The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.

We want to know the concentration of this diluted solution.

As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:

n_1= n_2  (1)

and we also know that:

n = M x V_2

If we replace this expression in (1) we have:

M_1 x V_1= M_2 x V_2

Where 1, would be the stock solution and 2, the solution we want to prepare.

So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is M_2:

4.93 x 210 =  620 xM_2

M_2 = 1.66 M

This is the concentration of the solution prepared.

Learn more about molarity here:

brainly.com/question/19517011

#SPJ1

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How many moles are in 1.5L of 0.40M Na2SO4?​
anzhelika [568]

Answer:

0.60 moles of Na₂SO₄

Explanation:

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A solution of Na₂SO₄ 0.40M contains 0.40 moles of solute (Na₂SO₄) per liter of solution.

As you have 1.5L of solution, moles of Na₂SO₄:

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