Respuesta :

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

[tex] Molarity = \frac{68.2 mol}{80*0.315 L} =2.71 \frac{mol}{L} Molarity = 2.71 M[/tex]

Answer: The molarity of solution is 2.7 M

Explanation:

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

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]

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:

[tex]\text{Molarity of solution}=\frac{68.2g\times 1000}{80g/mol\times 315mL}\\\\\text{Molarity of solution}=2.7M[/tex]

Hence, the molarity of solution is 2.7 M