Asked by rohan
4 gm of N aOH are present in 0.1dm3
solution have specific gravity 1.038 gm/ml. calculate:
a. Molality of N aOH solution;
b. Molarity of N aOH solution.
solution have specific gravity 1.038 gm/ml. calculate:
a. Molality of N aOH solution;
b. Molarity of N aOH solution.
Answers
Answered by
bobpursley
Molarity=moles/literssolution
= (4/40)/.1
Molality=moles/kgsolution
= (4/40)/(density*volume)
= (4/40)/(1.038g/ml*100ml)
= (4/40)/.1
Molality=moles/kgsolution
= (4/40)/(density*volume)
= (4/40)/(1.038g/ml*100ml)
Answered by
bobpursley
Molarity=moles/literssolution
= (4/40)/.1
Molality
= You have to know the volume of solvent. so start with the density of the solution and mass.
massolution=density*volume=1.038g/ml*100ml=103.8 grams
but you know you added 4g of NaOH, so the mass of the solvent must have been 99.2grams
Molality=molessolute/kgsolvent
= (4/40)/(.0992)
= (4/40)/.1
Molality
= You have to know the volume of solvent. so start with the density of the solution and mass.
massolution=density*volume=1.038g/ml*100ml=103.8 grams
but you know you added 4g of NaOH, so the mass of the solvent must have been 99.2grams
Molality=molessolute/kgsolvent
= (4/40)/(.0992)
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