Question

What is the mass in grams of na 2 ​ co 3 ​ (s) (molar mass = 105.98 g/mol) that, when dissolved in 125 ml of water, will make a 0.15 m aqueous solution?

Answer 1

`m_(Na_2CO_3)=2.0gNa_2CO_3`

Step-by-step explanation:

Hello,

In this case, we consider the equation defining molarity, in order to compute the mass that is present into 125 mL of an aqueous 0.15M of sodium carbonate (solute), Na₂CO₃ as shown below:

`M=(n_(solute))/(V_(solution))`

Now, since the unknown is the mass which comes from the moles, by solving for it and subsequently using its molar mass, one obtains:

`m_(Na_2CO_3)=125mL*(1L)/(1000mL)*0.15\frac{mol{Na_2CO_3}}{L}*\frac{105.98gNa_2CO_3}{1mol{Na_2CO_3}}   \\m_(Na_2CO_3)=2.0gNa_2CO_3`

Best regards.

Answer 2

Molarity is defined as the number of moles of solute dissolved in 1 L of solution.
molarity of sodium carbonate is 0.15 M
this means that 0.15 moles of sodium carbonate are dissolved in 1 L of solution
if 1 L contains - 0.15 mol
then 125 mL should contain - 0.15 mol/L x 0.125 L = 0.019 mol
mass of sodium carbonate in 125 mL - 0.019 mol x 105.98 g/mol = 2.0 g
2.0 g of sodium carbonate is required to make a 0.15 M solution