by 1°C (Kb of water = 0.512°C/m)?"

Answering this question requires the equation for boiling point elevation: ΔTb =iKbCm, where i is the van't Hoff factor, Kb is the boiling point elevation constant, and Cm is the solution molality. Since NaCl is an ionic species that dissociates into two ions (Na+ and Cl-) per mole, i = 2. We are given Kb = 0.512°C/m, and we are asked to find the amount of NaCl to raise the temperature by 1°C (ΔTb = 1°C).

Let's plug these in to solve for Cm:

1°C = (2)(0.512°C/m)(C m)

Cm = 0.977 m

This tells us the

*molality*of NaCl, whereas we want to know how many

*grams*are needed. Molality is equal to moles solute divided by liters solvent:

0.977 m = (X mol) / (1.0 kg)

X mol = 0.977 mol

Therefore, we need 0.977 moles of NaCl. To find the mass required, we can find its molecular mass based on the periodic table (58.4 g/mol) and solve for mass:

58.4 g/mol = (X g) / (0.977 mol)

X = 57 g NaCl