If 5. 0 grams of kcl is dissolved in 500 ml of water, what is the concentration of the resulting solution?
1 answer:
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Solution :
It is given that :
Weight of the antacid tablet = 5.4630 g
4.3620 gram of antacid is crushed and is added to the stomach acid of 200 mL and is reacted.
25 mL of the stomach acid that is partially neutralized required 13.6 mL of NaOH to be titrated for a red end point.
27.7 mL of solution is equivalent to
of the original stomach acid. Therefore, 13.6 mL of NaOH will take x
= 12.27 ml of the original stomach acid.
Answer : The mass of required is, 166.4 grams.
Explanation :
First we have to calculate the moles of nitrogen gas.
Using ideal gas equation:
where,
P = Pressure of gas = 1.00 atm
V = Volume of gas = 113 L
n = number of moles = ?
R = Gas constant =
T = Temperature of gas =
Putting values in above equation, we get:
Now we have to calculate the moles of sodium azide.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 3 mole of produced from 2 mole of
So, 3.84 moles of produced from
moles of
Now we have to calculate the mass of
Molar mass of = 65 g/mole
Therefore, the mass of required is, 166.4 grams.