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How many moles of NaOH are present in 294.2 mL of 1.747 M NaOH?
Solution
1.747 M NaOH is 1.747 mol NaOH / 1 L of solution
now
volume of the solution = 294.2 ml = 294.2 ml x 1 L / 1000 ml
volume of the solution = 0.2942 L
now
moles of NaOH = volume of solution x molarity
moles of NaOH = 0.2942 L x 1.747 mol / L
moles of NaOH = 0.513967 mol
so
there are 0.514 moles of NaOH

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How many moles of NaOH are present in 294.2 mL of 1.747 M NaOH S.pdf

  • 1. How many moles of NaOH are present in 294.2 mL of 1.747 M NaOH? Solution 1.747 M NaOH is 1.747 mol NaOH / 1 L of solution now volume of the solution = 294.2 ml = 294.2 ml x 1 L / 1000 ml volume of the solution = 0.2942 L now moles of NaOH = volume of solution x molarity moles of NaOH = 0.2942 L x 1.747 mol / L moles of NaOH = 0.513967 mol so there are 0.514 moles of NaOH