Answer: 581 gmol
0.581 kmol
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 
of particles.
To calculate the moles, we use the equation:
1. The conversion for mol to gmol
1 mol = 1 gmol
581 mol= 
2. The conversion for mol to kmol
1 mol = 0.001 kmol
581 mol= 
3. The conversion for mol to lbmol
1 mol = 
581 mol= 
25 drops of acid is required to neutralize the 50.0 ml of 0.010m of NaOH in the experiment.
The equation of the reaction is;
NaOH(aq) + HCl(aq) ---------> NaCl(aq) + H2O(l)
We can use the titration formula;
CAVA/CBVB = NA/NB
CA= concentration of acid
VA = volume of acid
CB = concentration of base
VB = volume of base
NA = number of moles of acid
NB = number of moles of base
CB = 0.010 M
VB = 50.0 ml
CA = 0.50 M
VA = ?
NA = 1
NB = 1
Substituting values;
CAVANB = CBVBNA
VA = 0.010 × 50.0 × 1/ 0.50 × 1
VA = 1 ml
Since the total volume of acid used is 1 ml and each drop contains 0.040 ml
The number of drops required is 1ml/0.040 ml = 25 drops
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