What is per cent by mass of na2co3 in a 1.0 molal aqueous solution ?

Hello!

We have the following data:

MM (Molar mass of Na2CO3) 
Na = 2*23 = 46 u
C = 1*12 = 12 u
O = 3*16 = 48 u
----------------------
MM (Molar mass of Na2CO3) = 46+12+48 = 106 g/mol
n (mol number) = 1 mol
m1 (mass of the solute) = ?









So if we have the dissolution in a liter (m2 - mass of solvent), that is 1000g , then the total mass of the solution (m) will be:







Now, let's find the percentage in mass (% m / m), let's see:









B) 9.6

I Hope this helps, greetings ... DexteR!

If a hydrogen atom is neutral, it must also have  one electron. an oxygen atom (o) has an atomic number of eight (8). this means a neutral oxygen atom has eight protons and eight electrons. the element actium (ac) has an atomic number of 89, so it has 89 protons and 89 electrons in a neutral atom