To calculate the volume of a 2.38 M sodium thiosulfate solution that contains 50.0 g of sodium thiosulfate (Na\(_2\)S\(_2\)O\(_3\)), we first need to determine the number of moles of sodium thiosulfate in 50.0 g.
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Calculate the molar mass of sodium thiosulfate (Na\(_2\)S\(_2\)O\(_3\)):
- Sodium (Na): 22.99 g/mol × 2 = 45.98 g
- Sulfur (S): 32.07 g/mol × 2 = 64.14 g
- Oxygen (O): 16.00 g/mol × 3 = 48.00 g
Now, add these together to find the molar mass of Na\(_2\)S\(_2\)O\(_3\): \[ \text{Molar mass} = 45.98 , \text{g} + 64.14 , \text{g} + 48.00 , \text{g} = 158.12 , \text{g/mol} \]
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Calculate the number of moles of Na\(_2\)S\(_2\)O\(_3\) in 50.0 g: \[ \text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{50.0 , \text{g}}{158.12 , \text{g/mol}} \approx 0.3164 , \text{mol} \]
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Use the molarity to find the volume: Molarity (M) is defined as the number of moles of solute per liter of solution: \[ M = \frac{\text{moles}}{\text{volume (L)}} \] Rearranging the equation to solve for volume gives: \[ \text{Volume (L)} = \frac{\text{moles}}{M} = \frac{0.3164 , \text{mol}}{2.38 , \text{mol/L}} \approx 0.1335 , \text{L} \]
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Convert volume from liters to milliliters: \[ \text{Volume (mL)} = 0.1335 , \text{L} \times 1000 , \text{mL/L} = 133.5 , \text{mL} \]
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Round to 3 significant digits: The final volume of the solution is approximately: \[ \boxed{134} , \text{mL} \]
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