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Ivenika [448]
3 years ago
12

A 7.73 mass % aqueous solution of ethylene glycol (HOCH2CH2OH) has a density of 1.19 g/mL. Calculate the molarity of the solutio

n. Give your answer to 2 decimal places.
Chemistry
1 answer:
butalik [34]3 years ago
8 0

Answer:

Molarity for the solution is 1.48 M

Explanation:

Molarity involves the moles of solute that are contained in 1L of solution. It is a sort of concentration. The most usual.

7.73 by mass involves the mass of solute that are contained in 100 g of solution, so this one of ethylene glycol contains 7.73 g of it.

Let's determine the moles of solute

7.73 g / molar mass of ethylene glycol = moles

7.73 g / 62 g/mol = 0.124 mol

If the mass of solution is 100 g, we can determine the volume with density.

Density = Mass / Volume

Volume = Mass / Density

Volume = 100 g / 1.19 g/mL = 84.03 mL

In conclusion, 0.124 moles are contained in 84.03 mL

Molarity is mol/L, so let's convert the volume in L

84.03 mL . 1L / 1000 mL = 0.08403 L

0.124 mol / 0.08403 L = 1.48 M

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Volume can be measured in:
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Calculate the mass of water produced when 1.36 g of butane reacts with excess oxygen.
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Answer:

Mass of water produced= 1.8 g

Explanation:

Given data:

Mass of water produced = ?

Mass of butane = 1.36 g

Mass of oxygen = excess

Solution:

Chemical equation:

2C₄H₁₀ +13 O₂       →   8CO₂ + 10H₂O

Number of moles of butane:

Number of moles = mass/molar mass

Number of moles = 1.36 g/ 58.12 g/mol

Number of moles = 0.02 mol

Now we will compare the moles of butane with water.

                 C₄H₁₀         :        H₂O

                  2              :           10

                 0.02          :         10/2×0.02 = 0.1 mol

Mass of water produced:

Mass = molar mass × molar mass

Mass = 0.1 mol × 18 g/mol

Mass = 1.8 g

6 0
3 years ago
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