Question #166064

Given 0.2 M solution of NaCl in water with a specific gravity of 1.35.

- How many grams of NaCl are present in 800 mL of solution?
- What is the molality of the solution?

Expert's answer

"Specific \\; gravity = \\frac{\u03c1_{NaCl}}{\u03c1_{water}} \\\\\n\n1.35 = \\frac{\u03c1_{NaCl}}{1} \\\\\n\n\u03c1_{NaCl} = 1.35 \\;g\/mL \\\\\n\n\u03c1 = \\frac{m}{V} \\\\\n\nm = \u03c1 \\times V \\\\\n\nm_{solute} = 1.35 \\times 800 = 1080 \\;g = 1.08 \\;kg"

Proportion:

0.2 mol â€“ 1000 mL

x mol â€“ 800 mL

"x =\\frac{0.2 \\times 800}{1000}=0.16 \\;mol \\\\\n\nm = n \\times M"

M(NaCl) = 58.44 g/mol

"m(NaCl) = 0.16 \\times 58.44 = 9.35 \\;g"

9.35 grams of NaCl are present in 800 mL of solution.

The formula for molality is

m = moles of solute / kilograms of solvent

"= \\frac{0.16 \\;mol}{1.08 \\;kg} \\\\\n\n= 0.148 \\;mol\/kg"

0.148 mol/kg is the molality of the solution.

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