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Sagot :
To determine the number of moles of \( CH_3OH \) (methanol) in \( 50.0 \, \text{mL} \) of a \( 0.400 \, M \) solution, we can follow these steps:
1. Convert the volume from milliliters to liters:
To perform this conversion, we use the fact that \( 1 \, \text{L} = 1000 \, \text{mL} \). Therefore, we convert \( 50.0 \, \text{mL} \) to liters:
[tex]\[ \text{Volume in liters} = \frac{50.0 \, \text{mL}}{1000} = 0.050 \, \text{L} \][/tex]
2. Use the molarity formula to find the number of moles:
Molarity (M) is defined as the number of moles of solute per liter of solution. The formula to calculate the number of moles (\( n \)) from molarity (M) and volume (V in liters) is:
[tex]\[ n = M \times V \][/tex]
Given that the concentration is \( 0.400 \, \text{M} \) and the volume in liters is \( 0.050 \, \text{L} \), we can now calculate the number of moles of \( CH_3OH \):
[tex]\[ n = 0.400 \, \text{M} \times 0.050 \, \text{L} = 0.020 \, \text{moles} \][/tex]
To summarize:
- The volume in liters is \( 0.050 \, \text{L} \).
- The number of moles of \( CH_3OH \) in the solution is \( 0.020 \) moles.
Thus, there are [tex]\( 0.020 \)[/tex] moles of [tex]\( CH_3OH \)[/tex] in [tex]\( 50.0 \, \text{mL} \)[/tex] of [tex]\( 0.400 \, M \)[/tex] [tex]\( CH_3OH \)[/tex].
1. Convert the volume from milliliters to liters:
To perform this conversion, we use the fact that \( 1 \, \text{L} = 1000 \, \text{mL} \). Therefore, we convert \( 50.0 \, \text{mL} \) to liters:
[tex]\[ \text{Volume in liters} = \frac{50.0 \, \text{mL}}{1000} = 0.050 \, \text{L} \][/tex]
2. Use the molarity formula to find the number of moles:
Molarity (M) is defined as the number of moles of solute per liter of solution. The formula to calculate the number of moles (\( n \)) from molarity (M) and volume (V in liters) is:
[tex]\[ n = M \times V \][/tex]
Given that the concentration is \( 0.400 \, \text{M} \) and the volume in liters is \( 0.050 \, \text{L} \), we can now calculate the number of moles of \( CH_3OH \):
[tex]\[ n = 0.400 \, \text{M} \times 0.050 \, \text{L} = 0.020 \, \text{moles} \][/tex]
To summarize:
- The volume in liters is \( 0.050 \, \text{L} \).
- The number of moles of \( CH_3OH \) in the solution is \( 0.020 \) moles.
Thus, there are [tex]\( 0.020 \)[/tex] moles of [tex]\( CH_3OH \)[/tex] in [tex]\( 50.0 \, \text{mL} \)[/tex] of [tex]\( 0.400 \, M \)[/tex] [tex]\( CH_3OH \)[/tex].
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