Question

B. if 6.73 g of na2co3 is dissolved in enough water to make 250. ml of solution, what is the molar concentration of sodium carbonate? what are the molar concentrations of na+ and co32- ions? (note 1 mol na2co3 = 2 mol of na+ = 1 mol co32-)

Answer 1

Answer:- $[Na_2CO_3]=0.254M$ , $[Na^+]=0.508 M$ , $[CO_3^2^-]=0.254M$

Solution:- We are asked to calculate the molarity of sodium carbonate solution as well as the sodium and carbonate ions.

Molarity is moles of solute per liter of solution. We have been given with 6.73 grams of sodium carbonate and the volume of solution is 250.mL.  Grams are converted to moles and mL are converted to L and finally the moles are divided by liters to get the molarity of sodium carbonate.

Molar mass of sodium carbonate is 105.99 gram per mol. The calculations for the molarity of sodium carbonate are shown below:

$\frac{6.73gNa_2CO_3}{250.mL}(\frac{1mol}{105.99g})(\frac{1000mL}{1L})$

= $0.254MNa_2CO_3$

So, molarity of sodium carbonate solution is 0.254 M.

sodium carbonate dissociate to give the ions as:

$Na_2CO_3(aq)\rightarrow 2Na^+(aq)+CO_3^2^-$

There is 1:2 mol ratio between sodium carbonate and sodium ion. So, the molarity of sodium ion will be two times of sodium carbonate molarity.

$[Na^+]=2(0.254M)$ = 0.508 M

There is 1:1 mol ratio between sodium carbonate and carbonate ion. So, the molarity of carbonate ion will be equal to the molarity of sodium carbonate.

$[CO_3^2^-]=0.254M$

Answer 2

Molarity is defined as the concentration of solute or a substance in given volume of a solution. The molar concentration of sodium carbonate will be  0.254 M.

Given that,

mass of sodium carbonate  = 6.73 g

volume of solution =250 mL

Molar mass of sodium carbonate = 105.99 g/mol

The molarity can be calculated as:

• $\text M &=\frac{n}{V}$

• $\frac{6.73 \;\text g}{250 \;\text{mL}}\times\frac{1 \text {mol}}{105.99\;\text g}\times\frac{1000 \;\text {ml}}{1 \;\text L}\\\\0.254 \text {M Na}_{2}\text{CO}_{3}$

Thus, molarity of sodium carbonate is 0.254 M.

The dissociation of sodium carbonate in the ions, we get:

• $\text{Na}_{2}\text{CO}_{3}\rightarrow 2\text{Na}^{+}+\text{CO}_{3}^{2-}$

From the above equation, it can be interpreted as there is 1:2 mol ratio between sodium carbonate and sodium ion. The molarity of sodium will be:

• $\text [{Na}^{+}] & = 2\times 0.254&=0.508 \text M$

Also, it is seen that there is 1:1 mol ratio between sodium carbonate and carbonate ion, therefore, its molarity will be:

• $\text[ {CO}_{3}^{2-}]&=0.254\;\text M$

Thus, the molarity of the sodium carbonate will be 0.254 M.

To know more about molarity, refer to the following link:

https://brainly.com/question/20878514?referrer=searchResults