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

Step-by-step explanation:

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

`Molarity=(n)/(V_s)`

where,

n = moles of solute

`V_s` = volume of solution in L

Molar mass of
`Na_2CO_3` =
`2* 22.99+1* 12.01+3* 16.00=105.99`

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

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

`Molarity=(0.462mol)/(2.00L)`

`Molarity=0.231M`

Therefore, the molarity of solution is 0.231 M