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jolli1 [7]
3 years ago
7

What is the molarity (M) of .5 liter of a solution that has 205 g of NaCl dissolved in it?

Chemistry
1 answer:
jeka943 years ago
8 0

Answer:

<h2>Molarity = 7 mol / L</h2>

Explanation:

Since the mass of NaCl and it's volume has been given we can find the molarity by using the formula

<h3>C =  \frac{m}{M \times v}</h3>

where

C is the molarity

m is the mass

M is the molar mass

v is the volume

From the question

v = 0.5 L

m = 205 g

We must first find the molar mass and then substitute the values into the above formula

M( Na) = 23 , M( Cl) = 35.5

Molar mass of NaCl = 23 + 35.5 =

58.5 g/mol

So the molarity of NaCl is

C =  \frac{205}{0.5  \times 58.5}  \\ C =  \frac{205}{29.25}

C = 7.00854

We have the final answer as

<h3>Molarity = 7 mol / L</h3>

Hope this helps you

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The correct answer is A. In the combined gas law, if the volume is decreased and the pressure is constant, then the temperature decreases.

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 <span>P1V1/ T1 = P2V1 /2 T2</span>

 <span>Cancelling terms,</span>

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 <span>Thus, the temperature decreased.</span>

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An element has the mass number 12 and atomic number 6. The number of neutron in it is?
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What are kind and number of atoms in ring portion in the haworth structure of glucose
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3 0
2 years ago
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A sample of nitrogen gas had a volume of 500. ml, a pressure in its closed container of 740 torr and a temperature 25 degrees c
Semenov [28]
Question:
              <span>A sample of nitrogen gas had a volume of 500mL, a pressure in its closed container of 740 torr and a temperature of 25°c. what was the volume of gas when the temperature was changed to 50°c and the new pressure was 760 torr?

Answer:

Data Given:
                   V</span>₁  =  500 mL

                   P₁  =  740 torr

                   T₁  =  25 °C + 273  =  298 K

                   V₂  =  ?

                   P₂  =  760 torr

                   T₂  =  50 °C + 273  =  323 K

Solution:
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                      P₁ V₁ / T₁  =  P₂ V₂ / T₂

Solving for V₂,

                      V₂  =  (P₁ V₁ T₂) ÷ (T₁ P₂)

Putting Values,
  
                      V₂  =  (740 torr × 500 mL × 323 K) ÷ (298 K × 760 torr)

                      V₂  =  527.68 mL
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