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Vsevolod [243]
3 years ago
11

Sodium carbonate (NaCO3) is sometimes used as a water-softening agent. Suppose that a worker prepares a 0.730 M solution of NaCO

3 and water. The volume of the solution is 1.421 liters. What is the mass of solute in the solution? Express your answer to three significant figures.
Chemistry
1 answer:
elena55 [62]3 years ago
4 0

Answer : The mass of solute in solution is 1.09\times 10^2g.

Solution : Given,

Molarity = 0.730 M

Volume of solution = 1.421 L

Molar mass of sodium carbonate = 105.98 g/mole

Formula used for Molarity :

Molarity=\frac{w}{M\times V}

where,

w = mass of solute

M = Molar mass of solute

V = volume of solution in liter

Sodium carbonate is solute and water is solvent.

Now put the given values in above formula, we get the mass of solute in solution.

0.730mole/L=\frac{w}{(105.98g/mole)\times (1.412L)}

By rearranging the terms, we get

m=109.247g=1.09\times 10^2g

Therefore, the mass of solute in solution is 1.09\times 10^2g.

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Explanation:

Based on the graph of US population and steel consumption, what could have led to the increase in steel consumption seen  on the graph is that whenever the population has increased, steel consumption has increased as well.

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What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 2.0 atm and the temperatu
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2) 2.066 atm.

Explanation:

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<em>PV = nRT.</em>

where, P is the pressure of the gas.

V is the volume of the container.

n is the no. of moles of the gas.

R is the general gas constant.

T is the temperature of the gas (K).

<em>1) What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 2.0 atm and the temperature is held constant?</em>

  • At constant T and at two different (P, and V):

<em>P₁V₁ = P₂V₂.</em>

P₁ = 2.0 atm, V₁ = 150.0 mL.

P₂ = ??? atm, V₂ = 50.0 mL.

<em>∴ P₂ = P₁V₁/V₂</em> = (2.0 atm)(150.0 mL)/(50.0 mL) = <em>6.0 atm.</em>

<em>2. A sample of a gas in a rigid container at 30.0°C and 2.00 atm has its temperature increased to 40.0°C. What will be the new pressure?</em>

<em></em>

  • Since the container is rigid, so it has constant V.
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<em>P₁/T₁ = P₂/T₂.</em>

P₁ = 2.0 atm, T₁ = 30.0°C + 273 = 303 K.

P₂ = ??? atm, T₂ = 40.0°C + 273 = 313 K.

<em>∴ P₂ = P₁T₂/T₁ </em>= (2.0 atm)(313.0 K)/(303.0 K) =<em> 2.066 atm.</em>

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