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Answer :
The amount of CH₃OH needed to make a 0.244 m solution in 400 g of water is 3.13g
we can use the formula:
molarity (M) = moles of solute (n) / volume of solution (V)
We know the molarity (0.244 M), the volume of solution (400 g of water), and we want to find the moles of solute. Rearranging the formula, we get:
moles of solute (n) = molarity (M) x volume of solution (V)
First, let's convert the volume of solution from grams of water to liters:
400 g water x 1 L / 1000 g water = 0.4 L
Now, we can plug in the values:
n = 0.244 M x 0.4 L = 0.0976 moles of CH₃OH
Finally, we can convert moles to grams using the molar mass ofCH₃OH:
0.0976 moles x 32.04 g/mol = 3.13 g
Therefore, the answer is (b) 0.313 g of CH₃OH is needed to make a 0.244 m solution in 400 g of water.
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