Answer:
0.595 M
Explanation:
The number of moles of water in 1L = 1000g/18g/mol = 55.6 moles of water.
Mole fraction = number of moles of KNO3/number of moles of KNO3 + number of moles of water
0.0194 = x/x + 55.6
0.0194(x + 55.6) = x
0.0194x + 1.08 = x
x - 0.0194x = 1.08
0.9806x= 1.08
x= 1.08/0.9806
x= 1.1 moles of KNO3
Mole fraction of water= 55.6/1.1 + 55.6 = 0.981
If
xA= mole fraction of solvent
xB= mole fraction of solute
nA= number of moles of solvent
nB = number of moles of solute
MA= molar mass of solvent
MB = molar mass of solute
d= density of solution
Molarity = xBd × 1000/xAMA ×xBMB
Molarity= 0.0194 × 1.0627 × 1000/0.981 × 18 × 0.0194×101
Molarity= 20.6/34.6
Molarity of KNO3= 0.595 M
To determine the number of moles of carbon dioxide that is produced, we need to know the reaction of the process. For the reaction of HCl and sodium carbonate, the balanced chemical equation would be expressed as:
2HCl + Na2CO3 = 2NaCl + H2O + CO2
From the initial amount given of sodium carbonate and the relation of the substances from the balanced reaction, we calculate the moles of carbon dioxide as follows:
0.2 moles Na2Co3 ( 1 mol CO2 / 1 mol Na2Co3 ) = 0.2 moles CO2
Therefore, the amount in moles of carbon dioxide that is produced from 0.2 moles sodium carbonate would be 0.2 moles as well.
Answer:
THE VOLUME OF 0.200M CALCIUM HYDROXIDE NEEDED TO NEUTRALIZE 35 mL of 0.050 M NITRIC ACID IS 43.75 mL.
Explanation:
Using
Ca VA / Cb Vb = Na / Nb
Ca = 0.0500 M
Va = 35 mL
Cb = 0.0200 M
Vb = unknown
Na = 2
Nb = 1
Equation for the reaction:
Ca(OH)2 + 2HNO3 --------> Ca(NO3)2 + 2H2O
So therefore, we make Vb the subject of the equation and solve for it
Vb = Ca Va Nb / Cb Na
Vb = 0.0500 * 35 * 1 / 0.0200 * 2
Vb = 1.75 / 0.04
Vb = 43.75 mL
The volume of 0.02M calcium hydroxide required to neutralize 35 mL of 0.05 M nitric acid is 43.75 mL
