Question

What is the molarity if 2.00 liters containing 49.0 grams of sodium carbonate [Na2CO3)?
(Molar mass of Na is 22.99 g/mol, C is 12.01 g/mol and O is 16.00 g/mol.)

Answer 1

Answer: The molarity of solution is 0.231 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Molar mass of $Na_2CO_3$ = $2\times 22.99+1\times 12.01+3\times 16.00=105.99$

moles of $Na_2CO_3$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{49.0g}{105.99g/mol}=0.462mol$

Now put all the given values in the formula of molality, we get

$Molarity=\frac{0.462mol}{2.00L}$

$Molarity=0.231M$

Therefore, the molarity of solution is 0.231 M