Answer : The vapor pressure of solution is 23.67 mmHg.
Solution:
As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.
The formula for relative lowering of vapor pressure will be,
where,
= vapor pressure of pure solvent (water) = 23.76 mmHg
= vapor pressure of solution= ?
= mass of solute (sucrose) = 12.25 g
= mass of solvent (water) = 176.3 g
= molar mass of solvent (water) = 18.02 g/mole
= molar mass of solute (sucrose) = 342.3 g/mole
Now put all the given values in this formula ,we get the vapor pressure of the solution.
Therefore, the vapor pressure of solution is 23.67 mmHg.
Answer:
I can provide a proper answer since there are no bonds specified.
Explanation:
Answer:
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<span> C12H4Cl6 + 23 O2 + 2 H2O → 24 CO2 + 12 HCl </span>
<span>(a) </span>
<span>(13.5 mol O2) x (2 mol H2O / 23 mol O2) = 1.17 mol H2O </span>
<span>(b) </span>
<span>(17.6 mol H2O) x (12 mol HCl / 2 mol H2O) x (36.4611 g HCl/mol) = 3850 g HCl </span>
<span>(c) </span>
<span>(90.4 g HCl) / (36.4611 g HCl/mol) x (24 mol CO2 / 12 mol HCl) = 4.96 mol CO2 </span>
<span>(d) </span>
<span>(106.01 g CO2) / (44.00964 g CO2/mol) x (2 mol C12H4Cl6 / 24 mol CO2) x (360.8782 g C12H4Cl6/mol) = </span>
<span>72.4 g C12H4Cl6 </span>
<span>(e) </span>
<span>(4.2 kg C12H4Cl6) / (360.8782 g C12H4Cl6/mol) x (12 mol HCl / 2 mol C12H4Cl6) x (36.4611 g HCl/mol) = </span>
<span>2.5 kg HCl</span>
