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marshall27 [118]
2 years ago
9

What is the molality of a solution containing 10.0 G of NA2 S04 and 1000.0 G of water

Chemistry
1 answer:
galina1969 [7]2 years ago
7 0

Explanation:

1 literThe total of water is equal to 1000.0 g of water

we need to find the molality of a solution containing 10.0 g of dissolved in  Na₂S0₄1000.0 g of water

1. For that find the molar mass

Na:  2 x 22.99= 45.98

S:  32.07

O:  4 x 16= 64

The total molar mass is 142.05

We have to find the number of moles, y

To find the number of moles divide 10.0g by 142.05 g/mol.

So the number of moles is 0.0704 moles.

For the molarity, you need the number of moles divided by the volume. So, 0.0704 mol/1 L.

The molarity would end up being 0.0704 M

The molality of a solution containing 10.0 g of Na2SO4 dissolved in 1000.0 g of water is  0.0704 Mliter

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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.

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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

\frac{d}{dt}[e^{\frac{1}{3}t}A] = 0.8e^{\frac{1}{3}t

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

∴ A(t) = 2.4 +Ce^{-\frac{1}{3}t

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

C= 12-2.4

C =9.6

Hence;

A(t) = 2.4 +9.6e^{-\frac{t}{3}}

A(0) = 2.4 +9.6e^{-\frac{10}{3}}

A(t) = 2.74

∴ the concentration at 10 minutes is ;

=  \frac{2.74}{6000}*100%

= 0.0456667 %

= 0.046% to three decimal places

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