A solution of naoh is 4g/l .What volume of hcl gas at stp will neutralize 50ml of slkaline solution
Answers
Answer:
A solution of NaOH is 4g/L .What volume of HCl gas at STP will neutralize 50ml of this alkaline solution?
The volume of HCl that neutralize the 50 ml of this NaOH solution is 112cm³
Explanation:
Let's calculate the moles of the NaOH from the volume given
Calculate mass of 50ml NaOH in 4g/L solution of NaOH
We have been given that the concentration is 4g/L, that means in 1 liter of this NaOH solution there is 4 g dissolved.
The volume given is 50 ml which 50 /1000 = 0.05L
Therefore if 1 L = 4g
Then 0.05 L = 4g × 0.05L / 1 L
= 0.2 g
The mass in 50 ml of this NaOH is 0.2g
Find moles of NaOH
Moles = mass /molar mass
Molar mass of NaOH = 40g/mol
Mass = 0.2g
Moles = 0.2g/ 40gmol⁻¹
= 0.005 moles
Use mole ratio from the equation of the neutralization reaction to calculate the moles of HCl gas used up:
The equation for the reaction:
HCl + NaOH = H₂O + NaCl
The mole ratio from the equation of HCl:NaOH is 1:1
Therefore the moles of HCl used up in the neutralization of the NaOH will also be 0.005 moles.
Find the volume of the HCl gas used using the Ideal gas Law:
Ideal gas law: 1 mole of any gas occupies 22.4 liters by volume at STP
Therefore if 1 mole = 22.4L
Then 0.005 moles = 22.4 L × 0.005 moles/ 1 mole
= 0.112 L
= 0.112 × 1000cm³
= 112 cm³
Therefore the volume of HCl used is 112 cm³
Explanation:
Option (2) 112 mL
Explanation:
The balanced chemical reaction is
NaOH + HCl → NaCl + H₂O
In 1 Litre, the NaoH = 4gm
or in 1000 ml, the NaOH = 4 gm
Therefore in 50 ml, NaoH = 0.2 gm
Moles of NaOH = Weight (gm)/Molecular Weight
= 0.2/40 = 0.005 mole
From the reation
1 mole of NaOH will neutralise 1 mole of HCl
Therefore 0.005 mole of NaCl will neutralise 0.005 mole of HCl
Volume of 1 mole of HCl is 22.4 litres at STP
Therefore volume of 0.005 moles of HCl is 0.005 x 22.4
= 0.112 Litre = 112 mL