Question #197761

Suppose you had 58.44 g of NaCl and you dissolved it in exactly 2.00 L of water. What would be the molality?

Expert's answer

The molality can be calculated using the following equation:

M = n(solute)/m(solvent)

where n - moles of solute, m mass of solvent (kg)

n(NaCl) = m(NaCl) / Mr(NaCl) = 58.44 g / 58.44 g/mol = 1 mol

From here, the molality of the solution is:

M(NaCl) = n(NaCl) / m(H_{2}O) = 1 mol / 2 kg = 0.5 mol/kg

**Answer: 0.5 mol/kg**

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