Question

If a 68.2g sample of ammonium nitrate is dissolved in enough water to make 315mL of solution, what will be the molarity

Answer 1

Molarity = mol solute/L solution

315 mL = 0.315 L

68.2 g NH4NO3

M(NH4NO3)=2M(N)+3N(O)+4M(H)=2*14.0+3*16.0+4*1.0= 28.0+48.0+4.0= =80 g/mol

number mol (NH4NO3) = 68.2 g*1mol/80 g =(68.2/80 ) mol


`Molarity = (68.2 mol)/(80*0.315 L) =2.71 (mol)/(L) Molarity = 2.71 M`

Answer 2

Answer: The molarity of solution is 2.7 M

Step-by-step explanation:

To calculate the molarity of solution, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}`

We are given:

Mass of solute (ammonium nitrate) = 68.2 g

Molar mass of ammonium nitrate = 80 g/mol

Volume of solution = 315 mL

Putting values in above equation, we get:

`\text{Molarity of solution}=(68.2g* 1000)/(80g/mol* 315mL)\\\\\text{Molarity of solution}=2.7M`

Hence, the molarity of solution is 2.7 M