Answer :
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 :
[tex]Molarity=\frac{\text{Moles of solute}\times 1000}{\text{volume of solution}}[/tex]
Now put all the given values in this formula, we get the molarity of solution.
[tex]Molarity=\frac{1mole\times 1000}{750ml}=1.33mole/L[/tex]
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.
[tex]\text{Moles of }NH_4Cl=\frac{\text{Mass of }NH_4Cl}{\text{Molar mass of }NH_4Cl}=\frac{0.54g}{53.491g/mole}=0.01moles[/tex]
Now we have to calculate the concentration of solution.
[tex]Concentration=\frac{\text{Moles of }NH_4Cl}{\text{Volume of solution}}=\frac{0.01moles}{250ml}=4\times 10^{-5}mole/ml=4\times 10^{-5}\times 10^3=0.04mole/L[/tex]
[tex](1L=1000ml)[/tex]
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.
[tex]\text{Moles of }Na_2PO_4=\frac{\text{Mass of }Na_2PO_4}{\text{Molar mass of }Na_2PO_4}=\frac{492g}{163.94g/mole}=3.00moles[/tex]
Now we have to calculate the concentration of solution.
[tex]Concentration=\frac{\text{Moles of }Na_2PO_4}{\text{Volume of solution}}=\frac{3.00moles}{500ml}=6.00\times 10^{-3}mole/ml=6.00\times 10^{-3}\times 10^3=6.00mole/L[/tex]
[tex](1L=1000ml)[/tex]
Therefore, the molarity of the solution is, 6.00 mole/L
