19. There is a 4% by mass NaoH solution, which has a density of 1.1glmu. a. What
ID: 930165 • Letter: 1
Question
19. There is a 4% by mass NaoH solution, which has a density of 1.1glmu. a. What is the molarity(2 points)? b. What is the molality(2 points)? c. What is the percent by mass 2 points)? d. What is the mole fraction(2 points)? e. What is the volume should be used to make su00L of NaoH solution with a pH of 11(2 points? f. What is the volume from solution (e) should be used to titrated o.1M of 25mLof acetic acid Ka 1.8x10 to reach equivalent point(2 points)? g. What is the pH after 300 ml of solution (e) mixed with o.1M of 25mL of acetic acid(2 points)? Kas 18x10Explanation / Answer
ANS 1)
Molar concentration, also called molarity, amount concentration or substance concentration, is a measure of the concentration of a solute in a solution, or of any chemical species in terms of amount of substance in a given volume.
1 molar = 1 M = 1 mole/litre
4% of NaOH = 4 gm NaOH per 100 ml
So
(x) (0.100 lit ) = 4.0 gm /39.9969
X = 4.0 / (0.100) (39.9969)
X = 1.0 molar
Molarity of the 4% NaOH solution is 1.0 molar
ANS 2)
Molality :
Molality means no. of moles of solute per 1000 g or per kg of solvent
Molar mass of NaOH = 23+1+16 = 40 g/mol.
So, 4% NaOH aq. means 40 g or 1 mole NaOH in 1000 ml (=1000 g or 1kg) water.
Molality = 1 mol solute / 1kg solvent
=1 mol/kg.Volume of solution
= mass of solution (40+1000 = 1040 g) /density of 1.1 g per cc = 945.45 cc, which has 1 mole solute in it. Therefore no. of moles of solute per 1000 cc or 1 litre solution = 1000 x1 / 945.45= 1.053 mole per litre.
ANS 3)
% of solution by mass is 4%
ANS 4)
Mole fraction of NaOH = No. of moles of NaOH / Total moles
= 1.0 / (1.0 + 1000 g /18 g per mole)
= 1.0 /(1 + 55.55) = 0.017 (it is unitless).
Mole fraction of water = 55.55 /56.55 = 0.982.
Also, the sum of mole fractions is always 1.00.
Rememer, molality is a mass/ mass unit. So, it the best unit because it is temperature independent whereas molarity is mass /volume unit which is dependent on temp
ANS 5)
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