A solution containing 12.0% sodium Hydroxide by mass has a density of 1.131 g/ml . What volume of this solution contains 5.00 mol of NaOH
Answers
Answered by
16
let's assume 12% implies 12gm of NaOH in 100gm of solution
density of solution = 1.131gm/ml
therefore volume of solution = weight of solution / density of solution =12/1.131= 10.61 ml
no of moles of NaOH = weight of NaOH / molar mass of NaOH = 12/40 =0.3
therefore 0.3 mol in 10.61 ml of solution
morality = no of moles / volume of solution = 0.3/10.61 = 0.028
now A/Q vol of solution having 5 mol = 5/0.028 = 178.57ml
hope my answer is right
if any queries please ask
density of solution = 1.131gm/ml
therefore volume of solution = weight of solution / density of solution =12/1.131= 10.61 ml
no of moles of NaOH = weight of NaOH / molar mass of NaOH = 12/40 =0.3
therefore 0.3 mol in 10.61 ml of solution
morality = no of moles / volume of solution = 0.3/10.61 = 0.028
now A/Q vol of solution having 5 mol = 5/0.028 = 178.57ml
hope my answer is right
if any queries please ask
Anonymous:
That Is the wrong answer
If u dont know then dont answer
its all right ...why to panic .I tried my best....I'll recheck my ans
ok
hey.....I took wt of soln to b 12 gm instead of 100 gm..is ans 1474.925ml.......... please check
nope
it should b .........whats the actual ans?
The answer 1474.92 is correct...
you stupid.....
Answered by
8
Answer:1.47L
Explanation:
12gm(12/40 moles) of NaOH present in 100gm soln.
So 1 mole is present in (100×40)/12 = 333.34gm
Thus 5moles are present in 333.34×5=1666.6g
Now, density=mass/volume
Here d=1.131g/ml=1131g/L
Thus volume = 1666.6/1131 = 1.47L
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