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Dear student,
Semimolar solution stands that the molarity of the given solution is ½ M or 0.5 M.
Since molarity is the number of moles of solute in 1 L of the solution hence 0.5 M means 0.5 moles of NaCl in 1 L or 1000 mL or 1000 cm3 of the solution.
 Density = 1.16 g/ cm3
Mass of solution =    


Mass of NaCl = 0.50 moles NaCl x (58.5 g NaCl / 1 mole NaCl) = 29.3 g NaCl
( Here molar mass of NaCl = 58.5 g/mol)
Mass of solvent in solution = 1160 g - 29.3 g = 1131 g H2
Molality NaCl = moles of NaCl / mass of solvent in kg
Molality = 0.50 moles / 1.131 kg = 0.44 molal NaCl

Answer:  0.44 m

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Dear student,

Please find the solution in the below image.
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