Given :
2.5 mole of Sulfuric acid 
.
To Find :
Mass of sodium hydroxide will completely neutralize 2.5 mol of sulfuric acid
Solution :
Let us assume volume of water be 1 L .
Now , we know , to neutralize 1 mole of sulfuric acid we need 2 moles of NaOH .
So , for 2.5 mole sulfuric acid required 5 mole of NaOH .
Moles of NaOH ,
Molecular mass of NaOH , M.M = 58.44 g/mol .
Mass of 5 moles of NaOH :
Hence , this is the required solution .
I only got 50 points (which is not 100). :-)
Look at the graph. At 80 °C, about 38 g of solute is able to dissolve, and that’s for ever 100 g of water. That means that for every 150 grams of water, 57 grams of solute can dissolve (38/2 = 19 + 38 = 57 g) at 80 °C. Since 57 g is greater than 55 g, all for he sodium chloride should dissolve in 150 g of water at 80 °C - you can put all of that into a “mathematical explanation”.
The answer is Solid.
Have a great day m8!
