A student performs two melting-point determinations on a crystalline product. In one determination, the capillary tube contains a sample about 1–2 mm in height and the melting range is found to be 141°– 142°C. In the other determination, the sample height is 4–5 mm and the melting range is found to be 141°–145°C. Explain the broader melting-point range observed for the second sample. The reported melt- ing point for the compound is 143°C.

the second sample has a broader range because it has more impurity by percentage

Explanation:

separation techniques

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

i think it's letter a.