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Bess [88]
2 years ago
7

Which intermolecular force is characteristic of compounds with low molar mass, which are liquids at room temperature and have re

latively high boiling points? Which intermolecular force is characteristic of compounds with low molar mass, which are liquids at room temperature and have relatively high boiling points? covalent bonds London forces dipole-dipole forces hydrogen bonds ionic bonds
Chemistry
1 answer:
eduard2 years ago
6 0

Answer:

Hydrogen bonds.

Explanation:

The intermolecular forces are the forces that are presented between molecules in a substance. As higher is the force, as closer are the molecules, and more difficult will be to separate them.

Because of that, the solids have stronger forces than the liquids, which have stronger forces than the gases. Also, as strong the force, as higher will be the boiling point.

The ionic bond is the strongest and it's presented at ionic compounds, which are solids at room temperature and have high boiling points.

The covalent bonds are presented in the molecules and are formed when atoms share pairs of electrons. Between these molecules, the forces can be London forces, dipole-dipole forces, or hydrogen bonds.

The London forces are the weaker, they are presented at the nonpolar molecules, so it's easy to boil it. For a compound with low molar mass and London forces, it'll probably be at the gas phase at room temperature.

The dipole-dipole forces are presented at the polar molecules, and it's strong than the London forces. And the hydrogen bonds are a specific type of dipole-dipole forces, which is stronger and is formed between hydrogen and F, O, or N.

Because hydrogen has a low molar mass (1 g/mol), the compounds formed by it intends to have a low molar mass. So, to be liquid at room temperature, low mass, and high boiling point, it's more probable that the compound has hydrogen bonds.

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maxonik [38]

The vapour pressure of the solution is 23.4 torr.

Use <em>Raoult’s Law</em> to calculate the vapour pressure:  

<em>p</em>₁ = χ₁<em>p</em>₁°  

where  

χ₁ = the mole fraction of the solvent  

<em>p</em>₁ and <em>p</em>₁° are the vapour pressures of the solution and of the pure solvent  

The formula for vapour pressure lowering Δ<em>p</em> is  

Δ<em>p</em> = <em>p</em>₁° - <em>p</em>₁  

Δ<em>p</em> = <em>p</em>₁° - χ₁<em>p</em>₁° = p₁°(1 – χ₁) = χ₂<em>p</em>₁°  

where χ₂ is the mole fraction of the solute.  

<em>Step 1</em>. Calculate the <em>mole fraction of glucose </em>

<em>n</em>₂ = 18.0 g glu × (1 moL glu/180.0 g glu) = 0.1000 mol glu  

<em>n</em>₁ = 95.0 g H_2O × (1 mol H_2O/18.02 g H_2O) = 5.272 mol H_2O  

χ₂ = <em>n</em>₂/(<em>n</em>₁ + n₂) = 0.1000/(0.1000 + 5.272) = 0.1000/5.372 = 0.018 62  

<em>Step 2</em>. Calculate the <em>vapour pressure lowering</em>  

Δ<em>p</em> = χ₂<em>p</em>₁° = 0.018 62 × 23.8 torr = 0.4430 torr  

<em>Step 3</em>. Calculate the <em>vapour pressure</em>  

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3 years ago
What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m
djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the  solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of  sucrose.

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544.75 g x 1 Kg/ 1000 g = 0.544 kg

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m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for  volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

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