4gm of NaOH can be neutralised by?
1) 100ml of 1N HCl
2) 200ml of N/2 H2SO4
3) 1000ml of N/10 KOH
4) 2000 ml of M/20 H3PO4
In this question, the options are given below and in those only 1 and 2 options are only correct.
THANK YOU.
To determine which option can neutralize 4g of NaOH, we need to calculate the number of moles of NaOH and then compare it to the number of moles of the respective reactant in each option.
Step 1: Calculate the number of moles of NaOH.
The molar mass of NaOH (sodium hydroxide) is:
Na: 22.99 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
The molar mass of NaOH is:
22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol
To calculate the number of moles, divide the mass of NaOH by its molar mass:
Number of moles = Mass / Molar mass
Number of moles = 4g / 40.00 g/mol
Number of moles = 0.1 mol
So, we have 0.1 moles of NaOH.
Step 2: Compare the number of moles of NaOH with each of the options.
Option 1: 100 ml of 1N HCl
To determine the number of moles of HCl, we multiply the volume (in liters) by the normality (N).
1N HCl means 1 mole of HCl per 1 liter of solution.
100 ml = 100/1000 L = 0.1 L
Number of moles of HCl = Volume (L) x Normality (N)
Number of moles of HCl = 0.1 L x 1 N = 0.1 mol
Since the number of moles of NaOH (0.1 mol) is equal to the number of moles of HCl in Option 1 (0.1 mol), Option 1 can neutralize 4g of NaOH.
Option 2: 200 ml of N/2 H2SO4
To determine the number of moles of H2SO4, we multiply the volume (in liters) by the normality (N).
N/2 H2SO4 means 1 mole of H2SO4 per 2 liters of solution.
200 ml = 200/1000 L = 0.2 L
Number of moles of H2SO4 = Volume (L) x Normality (N)
Number of moles of H2SO4 = 0.2 L x (1/2) N = 0.1 mol
Since the number of moles of NaOH (0.1 mol) is equal to the number of moles of H2SO4 in Option 2 (0.1 mol), Option 2 can neutralize 4g of NaOH.
Option 3: 1000 ml of N/10 KOH
To determine the number of moles of KOH, we multiply the volume (in liters) by the normality (N).
N/10 KOH means 1 mole of KOH per 10 liters of solution.
1000 ml = 1000/1000 L = 1 L
Number of moles of KOH = Volume (L) x Normality (N)
Number of moles of KOH = 1 L x (1/10) N = 0.1 mol
Since the number of moles of NaOH (0.1 mol) is equal to the number of moles of KOH in Option 3 (0.1 mol), Option 3 can neutralize 4g of NaOH.
Option 4: 2000 ml of M/20 H3PO4
To determine the number of moles of H3PO4, we multiply the volume (in liters) by the molarity (M).
M/20 H3PO4 means 1 mole of H3PO4 per 20 liters of solution.
2000 ml = 2000/1000 L = 2 L
Number of moles of H3PO4 = Volume (L) x Molarity (M)
Number of moles of H3PO4 = 2 L x (1/20) M = 0.1 mol
Since the number of moles of NaOH (0.1 mol) is equal to the number of moles of H3PO4 in Option 4 (0.1 mol), Option 4 can neutralize 4g of NaOH.
Therefore, all four options can neutralize 4g of NaOH.
4g NaOH = 0.10 eqv. Wt.
If Eqv Wt Acid = Eqv Wt Base then => Neutralization
Choices 1,2 & 4 are all 0.10 eqv wt.
Therefore, 1,2 & 4 will neutralize 0.10 eqv wt NaOH.
Eqv Wt = Normality x Volume in Liters
Choice 3 is also a strong alkali base. Base + Base => NO Rxn.