Mass of water produced : 0.146 g
<h3>Further explanation</h3>
Given
33.2 mL of 0.245 M lithium hydroxide
Required
mass of water
Solution
Reaction
HNO₃ (aq) + LiOH (aq) → H₂O (l) + LiNO₃ (aq)
mol LiOH :
= M x V
= 0.245 x 33.2 ml
= 8.134 mmol
From the equation, the mol ratio of HNO₃ : H₂O = 1 : 1, so mol H₂O = 8.134 mmol
mass H₂O :
= mol x MW
= 8.134 x 10⁻³ mol x 18 g/mol
= 0.146 g
This is an application of Boyle's law:
P₁V₁ = P₂V₂. we don't have to convert volume and pressure to standard forms. we can even use the pressure with mmHg
1 atm = 760 mmHg
V₂ = P₁V₁ / P₂ = 745 x 500 / 760 = 490 ml
Note that here we assume constant temperature
Metals on the left side, metalloids on the staircase, nonmetals on right side