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Georgia [21]
3 years ago
8

a laboratory procedure calls for making 510.0 mL of a 1.6 M KNO3 solution. How much KNO3 in grams is needed

Chemistry
1 answer:
barxatty [35]3 years ago
4 0

Answer:

82.416 g of KNO ₃  is needed to produce 510.0 mL of a 1.6 M KNO ₃ solution.

Explanation:

Since molarity is the number of moles of solute that are dissolved in a given volume, calculated by dividing the moles of solute by the volume of the solution, the following rule of three can be applied: if in 1 L (1,000 mL) of KNO₃ there are 1.6 moles of the compound present, in 510 mL how many moles will there be?

moles=\frac{510 mL*1.6 moles}{1000 mL}

moles= 0.816

Being the molar mass of the elements:

  • K: 39 g/mole
  • N: 14 g/mole
  • O: 16 g/mole

So the molar mass of the compound KNO₃ is:

KNO₃= 39 g/mole + 14 g/mole + 3*16 g/mole= 101 g/mole

Now I can apply the following rule of three: if in 1 mole of KNO₃ there are 101 g, in 0.816 moles how much mass is there?

mass=\frac{0.816 moles*101 grams}{1 mole}

mass= 82.416 grams

<u><em>82.416 g of KNO ₃  is needed to produce 510.0 mL of a 1.6 M KNO ₃ solution.</em></u>

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Explanation:

Step 1: Given data

  • Initial temperature (T₁): 30 °C
  • Initial volume (V₁): 3.25 L
  • Final temperature (T₂): -10 °C
  • Final volume (V₂): ?

Step 2: Convert the temperatures to Kelvin

We will use the following expression.

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T₁: K = 30°C + 273.15 = 303 K

T₂: K = -10°C + 273.15 = 263 K

Step 3: Calculate the final volume of the balloon

Assuming constant pressure and ideal behavior, we can calculate the final volume using Charles' law. Since the temperature is smaller, the volume must be smaller as well.

V₁/T₁ = V₂/T₂

V₂ = V₁ × T₂/T₁

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Instead of using ratios for back titrations we can also use molarities, if our solutions are standardized. A 0.196 g sample of a
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The total moles of HCl is:

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n_{HCl,excess} =n_{NaOH} =M_{NaOH} *V_{NaOH} =0.132*0.01105=1.46x10^{-3} moles

The difference between the total and excess of HCl, it can be know the moles that reacts with the aluminum hydroxide, is:

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