the volume of nh3 at stp needed to pass into 30 ml of 1N h2so4 solution to bring down its strength to m/10
Meq of original H2SO4=30×1=30
Meq of H2SO4 after passing NH3=30×0.2=6
Meq of H2SO4 lost =30−6=24
Meq of NH3 passed=Meq of H2SO4 lost
w17×1000=24
WNH3=0.408g
∴ Volume of NH3 at STP=22.4×0.40817
⇒0.5376l
⇒537.6ml
To find the volume of NH3 at STP required to dilute 30 mL of 1N H2SO4 solution to m/10, we need to use the concept of normality and the equation for dilution.
1. Calculate the number of moles of H2SO4 in the given solution:
N = Molarity of H2SO4 solution = 1N
Volume = 30 mL = 0.03 L
Number of moles of H2SO4 = N * Volume (in L)
Number of moles of H2SO4 = 1N * 0.03 L = 0.03 moles
2. Define the new normality (m/10) you want to achieve:
New normality = m/10 = 0.1N
3. The balanced equation for the reaction between H2SO4 and NH3 is:
H2SO4 + 2NH3 --> (NH4)2SO4
From the balanced equation, we can see that 1 mole of H2SO4 reacts with 2 moles of NH3.
4. Calculate the number of moles of NH3 required for the reaction:
Moles of NH3 = (Number of moles of H2SO4) / (2 moles)
Moles of NH3 = 0.03 moles / 2 = 0.015 moles
5. Now, we have the number of moles of NH3 needed. We can calculate the volume of NH3 at STP using the ideal gas law:
Volume of NH3 = (Number of moles of NH3) * (22.4 L/mol)
Volume of NH3 = 0.015 moles * 22.4 L/mol = 0.336 L
Therefore, the volume of NH3 at STP needed to pass into 30 mL of 1N H2SO4 solution to bring down its strength to m/10 is approximately 0.336 Liters or 336 milliliters.
To determine the volume of NH3 at STP needed to dilute the 1N H2SO4 solution to m/10, we can use the concept of normality and the balanced chemical equation between NH3 and H2SO4.
First, let's understand the given information:
1. The initial solution is 1N H2SO4. This means that 1 mole of H2SO4 is present in 1 liter of the solution. So, we have a 1:1 mole ratio between H2SO4 and H+ ions.
2. We want to dilute the solution to m/10. This means that the final concentration of H+ ions should be one-tenth of the original concentration. Since the acidic component of H2SO4 is H+, we need to dilute the H+ ions in this case.
Now, let's proceed with the calculations step by step:
Step 1: Determine the number of moles of H+ ions in the initial 1N H2SO4 solution.
- 1N H2SO4 means 1 mole of H2SO4 in 1 liter of the solution.
- Since the ratio between H2SO4 and H+ ions is 1:2, we have 2 moles of H+ ions in 1 mole of H2SO4.
- Therefore, the number of moles of H+ ions in the initial solution = 2 moles/Liter.
Step 2: Calculate the final concentration of H+ ions needed for m/10 dilution.
- We want to reduce the concentration of H+ ions to 1/10th of the initial concentration.
- Final concentration of H+ ions = (1/10) * (2 moles/Liter).
- Final concentration of H+ ions = 0.2 moles/Liter.
Step 3: Determine the volume of NH3 required to neutralize the excess H+ ions and reduce the concentration to 0.2 moles/Liter.
- From the balanced chemical equation between NH3 and H2SO4, we know that the mole ratio between NH3 and H+ ions is 1:2.
- Also, from the volume given in the question, we have 30 mL = 0.03 Liters.
- Volume of NH3 required = (0.2 moles/Liter) * (0.03 Liters) / (2 moles/Liter).
- Volume of NH3 required = 0.003 Liters = 3 mL.
Therefore, you would need 3 mL of NH3 at STP to pass into the 30 mL of 1N H2SO4 solution to bring down its strength to m/10.