Explanation:
Gay-Lussac's law states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas when the volume is kept constant. Mathematically, it can be written as: {\displaystyle {\frac {P}{T}}=k}. It is a special case of the ideal gas law.
The molarity and normality of 5.7 g of Ca(OH)2 in 450ml 0f solution is calculated as follows
molarity = moles/volume in liters
moles =mass/molar mass
= 5.7g/74g/mol = 0.077moles
molarity = 0.077/450 x1000= 0.17M
Normality = equivalent point x molarity
equivalent point of Ca(OH)2 is 2 since it has two Hydrogen atom
normality is therefore = 0.17 x2 = 0.34 N
Answer:
18 g
Explanation:
We'll begin by converting 500 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
500 mL = 500 mL × 1 L / 1000 mL
500 mL = 0.5 L
Next, we shall determine the number of mole of the glucose, C₆H₁₂O₆ in the solution. This can be obtained as follow:
Volume = 0.5 L
Molarity = 0.2 M
Mole of C₆H₁₂O₆ =?
Molarity = mole / Volume
0.2 = Mole of C₆H₁₂O₆ / 0.5
Cross multiply
Mole of C₆H₁₂O₆ = 0.2 × 0.5
Mole of C₆H₁₂O₆ = 0.1 mole
Finally, we shall determine the mass of 0.1 mole of C₆H₁₂O₆. This can be obtained as follow:
Mole of C₆H₁₂O₆ = 0.1 mole
Molar mass of C₆H₁₂O₆ = (12×6) + (1×12) + (16×6)
= 72 + 12 + 96
= 180 g/mol
Mass of C₆H₁₂O₆ =?
Mass = mole × molar mass
Mass of C₆H₁₂O₆ = 0.1 × 180
Mass of C₆H₁₂O₆ = 18 g
Thus, 18 g of glucose, C₆H₁₂O₆ is needed to prepare the solution.
