A student dissolves 106 g of #Na_2CO_3# in enough water to make a 6.00 L solution. What is the molarity of the solution?

Question

Caleb Adams · Accepted Answer

#0.1667M#

Take the formula mass of #Na_2CO_3#
Find the number of moles by dividing the amount of #Na_2CO_3# as provided in the problem by its formulas mass;
Compute the Molarity (M) using the formula;
M=number of mol/L solution
Make sure to cancel out units leaving only the desired unit
Per calculation, M=0.1667 mol/L

Olivia Anton · Answer

undefined To find the molarity (M) of the solution, use the formula:

$$ M = \frac{{\text{{moles of solute}}}}{{\text{{volume of solution in liters}}}} $$

First, calculate the moles of Na₂CO₃:

$$ \text{{Molar mass of Na₂CO₃}} = 2 \times \text{{molar mass of Na}} + 1 \times \text{{molar mass of C}} + 3 \times \text{{molar mass of O}} $$
$$ = 2 \times 22.99 \, \text{g/mol} + 1 \times 12.01 \, \text{g/mol} + 3 \times 16.00 \, \text{g/mol} $$
$$ = 2 \times 22.99 \, \text{g/mol} + 12.01 \, \text{g/mol} + 48.00 \, \text{g/mol} $$
$$ = 45.98 \, \text{g/mol} + 12.01 \, \text{g/mol} + 48.00 \, \text{g/mol} $$
$$ = 105.99 \, \text{g/mol} $$

$$ \text{{moles of Na}}_2\text{{CO}}_3 = \frac{{106 \, \text{g}}}{{105.99 \, \text{g/mol}}} $$

$$ \text{{moles of Na}}_2\text{{CO}}_3 ≈ 1.0 \, \text{mol} $$

Now, plug in the values into the formula:

$$ M = \frac{{1.0 \, \text{mol}}}{{6.00 \, \text{L}}} $$

$$ M ≈ 0.17 \, \text{M} $$

So, the molarity of the solution is approximately $0.17 \, \text{M}$.

Wyatt Atkins · Answer

undefined The molarity of the solution is approximately 3.51 M (molar) when 106 grams of Na2CO3 is dissolved in enough water to make a 6.00 L solution.