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Sagot :
Answer:
[tex]\boxed {\boxed {\sf 32 \ g \ NaHCO_3}}[/tex]
Explanation:
We are asked to find how much sodium bicarbonate (NaHCO₃) to weigh out.
1. Moles of Solute
We are given the molarity and the volume of solution. Molarity is a measure of concentration in moles per liter. The formula for molarity is:
[tex]molarity= \frac{moles \ of \ solute}{liters \ of \ solution}[/tex]
The molarity of the solution is 0.45 M or 0.45 moles of NaHCO₃ per liter. There are 850.0 milliliters of solution. We must convert milliliters to liters. Remember that 1 liter contains 1000 milliliters.
- [tex]\frac {1 \ L}{ 1000 \ mL}[/tex]
- [tex]850.0 \ mL * \frac {1 \ L}{ 1000 \ mL}[/tex]
- [tex]850.0 * \frac {1 \ L}{ 1000 }= \frac {850.0 }{1000} \ L =0.85 \ L[/tex]
Now we know the molarity and liters of solution, but the moles of solute is still unknown.
- molarity= 0.45 mol NaHCO₃ / L
- moles of solute = x
- liters of solution = 0.85 L
Substitute these values into the formula.
[tex]0.45 \ mol \ NaHCO_3 / L = \frac{ x}{0.85 \ L}[/tex]
We are solving for x, so we must isolate the variable. It is being divided by 0.85 liters. The inverse operation of division is multiplication, so we multiply both sides of the equation by 0.85 L.
[tex]0.85 \ L *0.45 \ mol \ NaHCO_3 / L = \frac{ x}{0.85 \ L} *0.85 \ L[/tex]
[tex]0.85 \ L *0.45 \ mol \ NaHCO_3 / L =x[/tex]
The units of liters cancel.
[tex]0.85 *0.45 \ mol \ NaHCO_3 =x[/tex]
[tex]0.3825 \ mol \ NaHCO_3=x[/tex]
2. Grams of Solute
Next we must convert moles to grams using the molar mass. This is the mass of 1 mole of a substance. The values are found on the Periodic Table because they are equal to the atomic masses but the units are grams per mole instead of atomic mass units. Look up molar masses of the individual elements in the compound NaHCO₃
- Na: 22.9897693 g/mol
- H: 1.008 g/mol
- C: 12.011 g/mol
- O: 15.999 g/mol
There is a subscript of 3 after O in the formula. There are 3 moles of oxygen in 1 mole of the compound, so we multiply oxygen's molar mass by 3 before adding the other molar masses.
- O₃= 47.997 g/mol
- NaHCO₃ = 22.9897693 + 1.008 + 12.011 + 47.997 =84.0057693 g/mol
Set up a ratio using the molar mass.
[tex]\frac {84.0057693 \ g \ NaHCO_3}{ 1 \ mol \ NaHCO_3}[/tex]
Multiply by the moles we calculated.
[tex]0.3825 \ mol \ NaHCO_3 *\frac {84.0057693 \ g \ NaHCO_3}{ 1 \ mol \ NaHCO_3}[/tex]
The units of moles of sodium bicarbonate cancel.
[tex]0.3825 *\frac {84.0057693 \ g \ NaHCO_3}{ 1}[/tex]
[tex]32.13220676 \ g \ NaHCO_3[/tex]
The original measurements have 4 and 2 significant figures. Our answer must have the least number of sig figs, or 2. For the number we found, that is the ones place. The 1 in the tenths place tells us to leave the 2 in the ones place.
[tex]32 \ g \ NaHCO_3[/tex]
We would need to weigh approximately 32 grams of sodium bicarbonate.
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