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grandymaker [24]
2 years ago
15

Which solution has the higher boiling point?

Chemistry
1 answer:
dedylja [7]2 years ago
7 0

NaCl has higher boiling point than C₂H₆O₂ in H₂O.

<h3>What is molality?</h3>
  • The number of moles of solute in a solution equal to 1 kg or 1000 g of solvent is referred to as its molality.
  • The definition of molarity, on the other hand, is based on a certain volume of solution.
  • Mol/kg is a typical molality measurement unit in chemistry.
<h3>Calculation of boiling point:</h3>

When the non-volatile solute is dissolved in a solvent, the boiling point rises along with the molality (concentration) of the solute.

Given,

Mass of  C₂H₆O₂ (solute) = 15.0 g

Mass of solvent (H₂O) = 0.50 kg

Molar mass of C₂H₆O₂ = 62 g/mol

Thus, the moles of solute = 15.0 g x 1.0 mol solute/62 g/mol

= 0.2419 mol

Therefore, molality(m) of C₂H₆O₂ (solute) = Amount of solute (mol)/ Mass of solvent

= 0.2419/0.50

= 0.4838 m

Similarly,

Given ,

Mass of  NaCl (solute) = 15 g

Mass of solvent (H₂O) = 0.50 kg

Molar mass of NaCl (solute) = 58.44 g/mol

Thus, the moles of NaCl (solute) = 15.0 g x 1.0 mol solute/58.44

= 0.2566 mol

Therefore, molality(m) of NaCl (solute) = Amount of solute (mol)/ Mass of solvent

= 0.2566 mol/0.50 kg  

= 0.51 m

Hence, the molality of NaCl (solute) is more than the molality of C₂H₆O₂(solute).

So, with an increase in the solute's concentration (molality), the boiling point rises.

Therefore, NaCl has higher boiling point than C₂H₆O₂ in H₂O.

Learn more about boiling point here:

brainly.com/question/24168079

#SPJ4

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Compound C with molar mass 205gmol-1 , contains 3.758g of carbon , 0.316g of hydrogen and 1.251g of oxygen. Determine the molecu
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Answer:

Molecular formula => C₁₂H₁₂O₃

Explanation:

From the question given above, the following data were obtained:

Molar mass of compound = 205gmol¯¹

Mass of Carbon (C) = 3.758 g

Mass of Hydrogen (H) = 0.316 g

Mass of Oxygen (O) = 1.251 g

Molecular formula =?

We'll begin by determining the empirical formula of the compound. This can be obtained as follow:

C = 3.758 g

H = 0.316 g

O = 1.251 g

Divide by their molar masses

C = 3.758 /12 = 0.313

H = 0.316 /1 = 0.316

O = 1.251 /16 = 0.078

Divide by the smallest.

C = 0.313 / 0.078 = 4

H = 0.316 / 0.078 = 4

O = 0.078 / 0.078 = 1

Thus, the empirical formula of the compound is C₄H₄O

Finally, we shall determine the molecular formula of the compound. This can be obtained as follow:

Molar mass of compound = 205gmol¯¹

Empirical formula => C₄H₄O

Molecular formula =>?

[C₄H₄O]ₙ = 205

[(12×4) + (4×1) +16]n = 205

[48 + 4 +16]n = 205

68n = 205

Divide both side by n

n = 205 / 68

n = 3

Molecular formula => [C₄H₄O]ₙ

Molecular formula => [C₄H₄O]₃

Molecular formula => C₁₂H₁₂O₃

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