1.prepare 2L of 0.150 deci normal aluminum chloride
2.saline solution contains 2grams of its major constituent solute in 1000ml of water.what is it's normality
To find the normality of a solution, we need to first understand the concept of normality. Normality is a measure of concentration that takes into account the chemical equation of the reaction. It is defined as the number of gram equivalents of solute per liter of solution.
Let's start by solving the first question:
1. Prepare 2 liters of 0.150 deci normal aluminum chloride.
To prepare the solution, we need to know the molar mass of aluminum chloride (AlCl₃). The molar mass of Al is 27 g/mol, and Cl is 35.5 g/mol (since chlorine is a diatomic molecule).
The deci normality (denoted as dN) of a solution represents that it is a tenth of a normal solution. Therefore, 0.150 deci normal is equivalent to 0.015 N (normality).
To calculate the grams of aluminum chloride needed, we can use the formula:
Grams of solute = (Normality * Equivalent weight * Volume of solution in liters)
The equivalent weight of aluminum chloride is the molecular weight divided by the number of moles of H+ or OH- produced in the reaction.
For aluminum chloride (AlCl₃), it dissociates into Al³⁺ and 3 Cl⁻ ions. This means the equivalent weight is (27 g/mol)/3 = 9 g/mol.
Now, substituting the values into the formula:
Grams of solute = (0.015 N) * (9 g/mol) * (2 L)
Grams of solute = 0.27 g
Therefore, to prepare 2 liters of 0.150 deci normal aluminum chloride, we need to dissolve 0.27 grams of aluminum chloride in the solution.
Now, let's move on to the second question:
2. A saline solution contains 2 grams of its major constituent solute in 1000 ml of water. What is its normality?
To find the normality, we need to know the molar mass and the valency of the solute.
The valency of sodium chloride (NaCl) is 1, and the molar mass is 23 g/mol for sodium and 35.5 g/mol for chlorine.
To calculate the normality, we can use the formula mentioned earlier.
Grams of solute = (Normality * Equivalent weight * Volume of solution in liters)
Since we have 2 grams of NaCl dissolved in 1000 ml of water, we need to convert the volume to liters by dividing it by 1000:
Volume of solution = 1000 ml / 1000 = 1 L
Substituting the values into the formula:
Grams of solute = (Normality) * (Equivalent weight) * (Volume of solution in liters)
2 g = (Normality) * (58.5 g/mol) * (1 L)
Solving for Normality:
Normality = (2 g) / (58.5 g/mol) = 0.034 N
Therefore, the normality of the saline solution is 0.034 N.