The molarity of a solution containing 5 gram of Sodium hydroxide (NaOH) in 500 millilitre solution is : (Take atomic weights of elements as : Na = 23, O = 16, H = 1)
The question asks us to find the molarity of a solution containing Sodium hydroxide (NaOH). Molarity is defined as the number of moles of solute dissolved per liter of solution. To calculate molarity, we need two things: the number of moles of NaOH and the volume of the solution in liters.
First, let's calculate the molar mass of Sodium hydroxide (NaOH). We are given the atomic weights:
The molar mass of NaOH is the sum of the atomic weights of its constituent atoms:
Molar mass of NaOH $ = $ Atomic weight of Na $ + $ Atomic weight of O $ + $ Atomic weight of H
Molar mass of NaOH $ = 23 + 16 + 1 = 40 $ g/mol
Next, we need to find the number of moles of NaOH present in the solution. We are given that the mass of NaOH is 5 grams.
The formula to calculate the number of moles is:
Number of moles $ = \frac{\text{Mass of substance}}{\text{Molar mass of substance}}$
Substituting the values:
Number of moles of NaOH $ = \frac{5 \text{ g}}{40 \text{ g/mol}}$
Number of moles of NaOH $ = 0.125 $ moles
The volume of the solution is given in millilitres (mL), but molarity requires the volume in litres (L). We need to convert 500 millilitres to litres.
There are 1000 millilitres in 1 litre.
Volume in litres $ = \frac{\text{Volume in millilitres}}{1000}$
Volume of solution $ = \frac{500 \text{ mL}}{1000 \text{ mL/L}} = 0.5 \text{ L}$
Now we have the number of moles of NaOH (0.125 mol) and the volume of the solution in liters (0.5 L). We can calculate the molarity using the formula:
Molarity (M) $ = \frac{\text{Number of moles of solute}}{\text{Volume of solution in litres}}$
Molarity $ = \frac{0.125 \text{ mol}}{0.5 \text{ L}}$
Molarity $ = 0.25 $ mol/L
So, the molarity of the solution is 0.25 mol/litre.
Let's check the options:
The calculated molarity matches option 4.
Which of the following correctly represents the number of atoms in one mole of CH3 OH?
A solution of Gallic acid in ethanol was made by dissolving 0.05 mg in 250 mL of ethanol(w/v). This solution in ppm would be ______________.