Answer :
The molarity of the solution made by dissolving 38.1-g sample of SrCl₂ in 112.5 mL of solution is B) 2.14 M.
To calculate the molarity of the SrCl₂ solution, you need to follow these steps:
1. Determine the molecular weight of SrCl₂. The atomic weights of Sr, Cl, and Cl are 87.62 g/mol, 35.45 g/mol, and 35.45 g/mol, respectively. So, the molecular weight of SrCl₂ is 87.62 + 35.45 + 35.45 = 158.52 g/mol.
2. Convert the mass of SrCl₂ into moles. You have a 38.1-g sample, so divide the mass by the molecular weight to find the moles: 38.1 g / 158.52 g/mol = 0.2403 mol.
3. Convert the volume of the solution into liters. You have 112.5 mL of solution, so divide by 1,000 to get 0.1125 L.
4. Calculate the molarity by dividing the moles of solute (SrCl₂) by the liters of solution: 0.2403 mol / 0.1125 L = 2.136 M.
The molarity of the SrCl2 solution is approximately 2.14 M, which corresponds to answer choice B.
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