Heat required to vaporize 1 mol of water from water at 100C to steam at 100C = 40.7 kJ
<span>1 mol of water weighs = 18.015g
</span>1.55 kg = <span>1550/18.015 mol = 86.03 mol
</span><span>Heat required to vaporize :
</span>= 86.03 mol x <span>40.7 kJ
</span>
= 3501.421 kJ
The balanced equation for the reaction between KOH and HBr is as follows;
KOH + HBr --> KBr + H₂O
stoichiometry of KOH to HBr is 1:1
number of KOH moles reacted - 0.25 mol/L x 0.015 L = 0.00375 mol
according to molar ration
number of KOH moles reacted = number of HBr moles reacted
number of HBr moles reacted - 0.00375 mol
if 12 mL of HBr contains - 0.00375 mol
then 1000 mL of HBr contains - 0.00375 mol / 12 mL x 1000 mL = 0.313 mol
therefore molarity of HBr is 0.313 M
Density = mass / volume
Density = 7.5 g / 5.0 cm3
Density = 1.5 g/cm3
