Answer:
At 0.58 L of 0.540 M NaOH solution contain 12.5 g NaOH.
Explanation:
Given data:
At volume = ?
Mass of NaOH = 12.5 g
Molarity of solution = 0.540 M
Solution:
First of all we will calculate the number of moles of sodium hydroxide.
Number of moles = mass/molar mass
Number of moles = 12.5 g / 40 g/mol
Number of moles = 0.3125 mol
Volume of NaOH:
Molarity = number of moles / volume in L
Now we will put the values.
0.540 M = 0.3125 mol / volume in L
volume in L = 0.3125 mol / 0.540 mol/L
volume in L = 0.58 L
63 mm multiply the length value by 10 bb ♡
The equation to be used are:
PM = ρRT
PV = nRT
where
P is pressure, M is molar mass, ρ is density, R is universal gas constant (8.314 J/mol·K), T is absolute temperature, V is volume and n is number of moles
The density of air at 23.5°C, from literature, is 1.19035 kg/m³. Its molar mass is 0.029 kg/mol.
PM = ρRT
P(0.029 kg/mol) = (1.19035 kg/m³)(8.314 J/mol·K)(23.5+273 K)
P = 101,183.9 Pa
n = 0.576 g * 1 kg/1000 g * 1 mol/0.029 kg = 0.019862 mol
(101,183.9 Pa)V = (0.019862 mol)(8.314 J/mol·K)(23.5+273 K)
Solving for V,
V = 4.839×10⁻⁴ m³
Since 1 m³ = 1000 L
V = 4.839×10⁻⁴ m³ * 1000
V = 0.484 L
The reaction is endothermic as energy is absorbed to break a bond.
