Answer to Question #166064 in Inorganic Chemistry for Wilhelm

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?
1
Expert's answer
2021-02-24T03:38:15-0500

Proportion:

0.2 mol – 1000 mL

x mol – 800 mL

M(NaCl) = 58.44 g/mol

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

The formula for molality is

m = moles of solute / kilograms of solvent

0.148 mol/kg is the molality of the solution.


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