By stoichiometry and assume
that:
CxH2xOy + zO2 -> xCO2
+ xH2O
<span>
CO2: 9.48/44 = 0.215 mmol
H2O: 3.87/18 = 0.215 mmol
mass of C = 0.215 * 12 = 2.58 mg
mass of H = 0.215 * 2 * 1 = 0.43 mg
mass of O in ethylbutyrate = 4.17 - 2.58 - 0.43 = 1.11 mg
So C/O = 2.58/1.11 ≈ 3 </span>
<span>
Thus we have C3H6O</span>
<span> </span>
This freezing point business is usually based on molality, that is moles solute per kilogram of solvent.
<span>molality = freezing point depression over Kf </span>
<span>In this case molality -10.3 degrees over -1.85 degrees Kf = 5.53 molal </span>
<span>This 5.53 molal solution is made up of l000 gms water and 5.53 moles glucose at 180 grams per mole for a total mass of 1997 grams </span>
<span>It volume would be l997 gms over 1.50 gms/ml or 1331 ml </span>
<span>We know that we have 5.53 moles of glucose dissolved in l331 ml of solution so now we can find how many moles of glucose in l000 ml or one liter of solution and this will be our Molarity </span>
<span>5.53 moles glucose over l331 ml = X moles glucose over l000 ml solution </span>
<span>cross multiply and solve for X moles glucose per liter solution </span>
<span>X = 4.15 moles glucose per liter = 4.15 Molar</span>
Answer:
Is this math? Cause as a fourth grader, I can do Algebra, but not this.
Explanation:
Answer:
i don get it
Explanation:i dont get it
