Answer:
0.0847M is molarity of sodium hydrogen citrate in the solution
Explanation:
The 2.0%(w/v) solution of sodium hydrogen citrate contains 2g of the solute in 100mL of solution. To find the molarity of the solution we need to convert the mass of solute to moles using molar mass and the mL of solution to Liters because molarity is the ratio between moles of sodium hydrogen citrate and liters of solution.
<em>Moles Na2C6H6O7:</em>
<em>Molar Mass:</em>
2Na: 2*22.99g/mol: 45.98g/mol
6C: 6*12.01g/mol: 72.01g/mol
6H: 6*1.008g/mol: 6.048g/mol
7O: 7*16g/mol: 112g/mol
45.98g/mol + 72.01g/mol + 6.048g/mol + 112g/mol = 236.038g/mol
Moles of 2g:
2g * (1mol / 236.038g) = <em>8.473x10⁻³ moles</em>
<em />
<em>Liters solution:</em>
100mL * (1L / 1000mL) = <em>0.100L</em>
<em>Molarity:</em>
8.473x10⁻³ moles / 0.100L =
<h3>0.0847M is molarity of sodium hydrogen citrate in the solution</h3>
