Question

Determine the molarity of a solution in mol/L that is prepared by completely dissolving 1.23 g of solid SrF2 (Mm=125.616g/m) in 417g of water.

Answer 1

Since we know that M=n/V, we use the initial data we have and then we proceed to calculate M:

$M=\cfrac{1.23\,g\,SrF_2}{417\,g\,H_2O}\times \cfrac{1\,mol\,SrF_2}{125.616\,g\,SrF_2}\times \cfrac{1\,g\,H_2O}{0.001\,L\,H_2O} \\ \text{ } \\ \implies M=0.0235 \,mol/L$

In conclusion, the molarity of the SrF₂ solution is M = 0.0235 mol/L.

Reference

• Chang, R., & Goldsby, K. A. (2010). Chemistry. Chemistry, 10th ed.; McGraw-Hill Education: New York, NY, USA.