<em>c</em> = 1.14 mol/L; <em>b</em> = 1.03 mol/kg
<em>Molar concentration
</em>
Assume you have 1 L solution.
Mass of solution = 1000 mL solution × (1.19 g solution/1 mL solution)
= 1190 g solution
Mass of NaHCO3 = 1190 g solution × (7.06 g NaHCO3/100 g solution)
= 84.01 g NaHCO3
Moles NaHCO3 = 84.01 g NaHCO3 × (1 mol NaHCO3/74.01 g NaHCO3)
= 1.14 mol NaHCO3
<em>c</em> = 1.14 mol/1 L = 1.14 mol/L
<em>Molal concentration</em>
Mass of water = 1190 g – 84.01 g = 1106 g = 1.106 kg
<em>b</em> = 1.14 mol/1.106 kg = 1.03 mol/kg
The balanced equation for the reaction is
CO(g) + 2H₂(g) ⇄ CH₃<span>OH(g)
Since given concentrations are at equilibrium state, the expression for the equilibrium constant, k can be written as
k = [</span>CH₃OH(g)] / [CO(g)] [H₂(g) ]²
By substitution,
k = 0.030 M / 0.020 M x (<span>0.072 M</span>)²
k = 289.35 M⁻²
Tell your teacher. They'll know what to do and it's best to report it to them.
