Question 9.11: What is the molarity (M) of 60.0 g of NaOH in 0.250 L of NaO...
What is the molarity (M) of 60.0 g of NaOH in 0.250 L of NaOH solution?
STEP 1 State the given and needed quantities.
| ANALYZE THE PROBLEM | Given | Need | Connect |
| 60.0 g of NaOH, 0.250 L of NaOH solution | molarity (mole/L) | molar mass of NaOH,
\frac{moles of solute }{liters of solution } |
To calculate the moles of NaOH, we need to write the equality and conversion factors for the molar mass of NaOH. Then the moles in 60.0 g of NaOH can be determined.
\begin{array}{r c}\boxed{\begin{matrix} 1 mole of NaOH = 40.00 g of NaOH \\\frac{40.00 g NaOH}{1 mole NaOH} \text{ and } \frac{1 mole NaOH}{ 40.00 g NaOH} \end{matrix}} \end{array}moles of NaOH = 60.0 \cancel{g NaOH} \times \frac{1 mole NaOH}{40.0 \cancel{g NaOH}}
= 1.50 moles of NaOHvolume of solution = 0.250 L of NaOH solution
STEP 2 Write the concentration expression.
Molarity (M) = \frac{moles of solute }{liters of solution }
STEP 3 Substitute solute and solution quantities into the expression and calculate.
M = \overset{Three SFs}{\underset{Three SFs}{\frac{1.50 moles NaOH}{0.250 L solution} }} = \overset{Three SFs}{\underset{Exact}{\frac{6.00 moles NaOH}{1 L solution} }} = \underset{Three Sfs }{6.00 M NaOH solution}Related Answered Questions
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