Question

1.Distinguish between a 1M solution and a 1m solution.

2. Calculate the molarity of 1.0 mol of KCl in 750 mL of solution.

3. What is the concentration (in M) of each of the following solutions?
a. 0.54g of ammonium chloride in 250 mL of solution b. 492g of sodium phosphate in 500 mL of solution

Answer 1

Explanation for part 1 :

1 M solution means that the one mole of solute present in one liter of solution and also known as molarity of the solution. The unit of molarity is mole/liter.

1 m solution means that the one mole of solute present in one kilogram of solvent and also known as molality of the solution. The unit of molality is mole/kilogram.

Explanation for part 2 :

Given : Moles of KCl (solute) = 1 mole

Volume of solution = 750 ml

Formula used for molarity :

`Molarity=\frac{\text{Moles of solute}* 1000}{\text{volume of solution}}`

Now put all the given values in this formula, we get the molarity of solution.

`Molarity=(1mole* 1000)/(750ml)=1.33mole/L`

Therefore, the molarity of the solution is, 1.33 mole/L

Explanation for part 3 :

Part (a) : Given,

Mass of ammonium chloride (solute) = 0.54 g

Volume of solution = 250 ml

Molar mass of ammonium chloride = 53.491 g/mole

First we have to calculate the moles of ammonium chloride.

`\text{Moles of }NH_4Cl=\frac{\text{Mass of }NH_4Cl}{\text{Molar mass of }NH_4Cl}=(0.54g)/(53.491g/mole)=0.01moles`

Now we have to calculate the concentration of solution.

`Concentration=\frac{\text{Moles of }NH_4Cl}{\text{Volume of solution}}=(0.01moles)/(250ml)=4* 10^(-5)mole/ml=4* 10^(-5)* 10^3=0.04mole/L`

`(1L=1000ml)`

Therefore, the molarity of the solution is, 0.04 mole/L

Part (b) : Given,

Mass of sodium phosphate (solute) = 492 g

Volume of solution = 500 ml

Molar mass of ammonium chloride = 163.94 g/mole

First we have to calculate the moles of sodium phosphate.

`\text{Moles of }Na_2PO_4=\frac{\text{Mass of }Na_2PO_4}{\text{Molar mass of }Na_2PO_4}=(492g)/(163.94g/mole)=3.00moles`

Now we have to calculate the concentration of solution.

`Concentration=\frac{\text{Moles of }Na_2PO_4}{\text{Volume of solution}}=(3.00moles)/(500ml)=6.00* 10^(-3)mole/ml=6.00* 10^(-3)* 10^3=6.00mole/L`

`(1L=1000ml)`

Therefore, the molarity of the solution is, 6.00 mole/L