How many grams of solute are contained in 64.3 mL of 0.0238M KOH?
Do I convert 64.3ml to L=.0643L
Multiiply .0643L and .0238M KOH=.00153
Multiply .00153 and 56.105 g/mol KOH=.0858g KOH
Procedure should be correct, did not check the calculations.
I agree with the procedure as well as the value and the number of significant figures.
Well, you did all the calculations correctly! But let me ask you something - why did the scarecrow win an award?
To calculate the number of grams of solute in 64.3 mL of 0.0238 M KOH, follow these steps:
Step 1: Convert mL to L.
- 64.3 mL * (1 L / 1000 mL) = 0.0643 L
Step 2: Multiply the volume in L and the molarity to find the number of moles of solute.
- 0.0643 L * 0.0238 M = 0.00153 moles
Step 3: Multiply the number of moles by the molar mass of KOH to find the number of grams of solute.
- 0.00153 moles * 56.105 g/mol = 0.0858 grams
Therefore, there are approximately 0.0858 grams of KOH in 64.3 mL of 0.0238 M KOH.
To determine the number of grams of solute in 64.3 mL of 0.0238 M KOH (potassium hydroxide), follow these steps:
1. Convert the milliliters (mL) to liters (L):
64.3 mL * (1 L / 1000 mL) = 0.0643 L
2. Multiply the volume in liters by the molarity of the solution (in moles per liter) to find the number of moles of solute:
0.0643 L * 0.0238 M = 0.0015314 moles KOH
3. Finally, calculate the number of grams by multiplying the number of moles by the molar mass of KOH:
0.0015314 moles KOH * 56.105 g/mol KOH = 0.0858 grams KOH
Therefore, the solution contains approximately 0.0858 grams of KOH.