How do you calculate the number of grams of solute in 250.0 mL of 0.179 M KOH?
We use the formula.....
Thus, (i) we determine the molar quantity.
Number of "Moles of KOH"="concentration"xx"volume"=0.179molL^-1xx250.0mLx10^-3LmL^-1=4.475x10^-2mol.
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To calculate the number of grams of solute (potassium hydroxide, KOH) in 250.0 mL of a 0.179 M (molar) solution, you can use the formula:
Number of grams of solute = Molarity × Volume × Molar mass
Given:
- Molarity (M) = 0.179 M
- Volume = 250.0 mL = 0.2500 L
- Molar mass of KOH (potassium hydroxide) = 56.11 g/mol (potassium: 39.10 g/mol, oxygen: 16.00 g/mol, hydrogen: 1.01 g/mol)
Now, plug in the values into the formula and calculate:
Number of grams of solute = 0.179 M × 0.2500 L × 56.11 g/mol
Number of grams of solute ≈ 2.507 g
Therefore, there are approximately 2.507 grams of potassium hydroxide (KOH) in 250.0 mL of a 0.179 M solution.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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