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3 years ago

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The density of an aqueous solution containing 25.0 percent of ethanol (C2H5OH) by mass is 0.950 g/mL. (a) Calculate the molality

of this solution. m (b) Calculate its molarity. M (c) What volume of the solution would contain 0.275 mole of ethanol

1 answer:

lora16 [44]3 years ago

6 0

Answer:

a. 7.24m

b. 5.15M

c. 53.4mL of the solution would contain this amount of ethanol.

Explanation:

Molality, m, is defined as the moles of solute (ethanol, in this case) per kg of solvent.

Molarity, M, are the moles of solute per kg of solvent

To solve this question we need to find the moles of solute in 100g of solution and the volume using its density as follows:

a. <em>Moles ethanol -Molar mass: 46.07g/mol-:</em>

25g ethanol * (1mol/46.07g) = <em>0.54265 moles ethanol</em>

<em>kg solvent:</em>

100g solution - 25g solute = 75g solvent * (1kg / 1000g) = 0.075kg

<em>Molality:</em>

0.54265 moles ethanol / 0.075kg = 7.24m

b. <em>Liters solution:</em>

100g solution * (1mL / 0.950g) = 105.3mL * (1L / 1000mL) = <em>0.1053L</em>

<em>Molarity:</em>

0.54265 moles ethanol / 0.1053L = 5.15M

c. 0.275 moles ethanol * (1L / 5.15moles Ethanol) = 0.0534L =

53.4mL of the solution would contain this amount of ethanol

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Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m

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Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

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Integration of the above linear equation =

$e^{\int\limits \frac {1}{3}dt } =$ $e^{\frac{1}{3}t$

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$e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A$ $= 0.8e^{\frac{1}{3}t$

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$Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C$

∴ $A(t) = 2.4 +Ce^{-\frac{1}{3}t$

Since A(0) = 12

Then;

$12 =2.4 + Ce^{-\frac{1}{3}}(0)$

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$A(t) = 2.74$

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= $\frac{2.74}{6000}*100$ %

= 0.0456667 %

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