From the chemical equation given:
H2SO4+2KOH--->K2SO4+2H2O
the two reactants, H2SO4 and KOH, are in 1:2 stoichiometric ratio.
No. of moles of KOH = 2* no. of moles of H2SO4
=2*0.1*0.033
The concentration of KOH = no. of moles / volume
=2*0.1*0.033/0.05
=0.132M
Answer: 1 molecule of
reacts with 1 molecule of
to give 2 molecules of HCl.
1 mole of
reacts with 1 mole of
to give 2 moles of HCl.
Explanation:
The given balanced reaction is:

On the particulate level :
1 molecule of
reacts with 1 molecule of
to give 2 molecules of HCl.
On molar level:
1 mole of
reacts with 1 mole of
to give 2 moles of HCl.
The new volume of the dilute solution is 0.33 L.
<u>Explanation:</u>
Using the law of Volumetric analysis, we can find the volume of the dilute solution from the stock solution or the concentrated solution of Calcium Chloride.
V1M1 = V2M2
V1 be the volume of the stock solution = 0.25 L
M1 being the molarity of the stock solution = 0.98 M
V2 be the volume of the dilute solution = ?
M2 being the molarity of the dilute solution = 0.74 M
We have to rearrange the above equation to find V2 as,
V2 = 
Now plugin the values as,
V2 = 
= 0.33 L
So the new volume of the dilute solution is 0.33 L.
Answer: 92mg of ethyl alcohol
2 milli molar of ethyl alcohol means 2 milli mole glucose for 1 liter of water. Since we want to make 1L of solution, then the amount of ethyl alcohol needed is: 2 milli mole/l x 1l= = 2 x 10^-3 mole
Ethyl alcohol molecular weight is 46g/mole, then 2 mili mole of glucose is= 2x 10^-3 mole x 46 grams/mole= 92 grams x 10^-3= 92mg
Answer: The freezing point of a 0.765 m solution of nitrobenzene in naphthalene is 
Explanation:
Depression in freezing point is given by:

= Depression in freezing point
i= vant hoff factor = 1 (for non electrolyte such as nitrobenzene)
= freezing point constant = 
m= molality = 0.765


Thus the freezing point of a 0.765 m solution of nitrobenzene in naphthalene is 