10 mL of 2 M NaOH solution is added to 20 mL of 1 M HCl solution kept in a beaker. Now, 10 mL of this mixture is poured into a volumetric flask of 100 mL containing 2 moles of HCl and made the volume upto the mark with distilled water. The solution in this flask is :
Show Hint
- 0.2 M NaCl solution
- 20 M HCl solution
- 10 M HCl solution
- Neutral solution
The Correct Option is B
Approach Solution - 1
To solve the question, let's analyze the steps involved in the whole process:
Determine the reaction between NaOH and HCl in the initial mixture:
- We have 10 mL of 2 M NaOH. Therefore, moles of NaOH = \(0.01 \, \text{L} \times 2 \, \text{mol/L} = 0.02 \, \text{mol}\).
- We have 20 mL of 1 M HCl. Therefore, moles of HCl = \(0.02 \, \text{L} \times 1 \, \text{mol/L} = 0.02 \, \text{mol}\).
- NaOH and HCl react in a 1:1 ratio: \(\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}\).
- Both reactants have the same number of moles (0.02 mol each), so they will completely neutralize each other, resulting in a neutral solution of NaCl with no excess HCl or NaOH.
Consider what happens when 10 mL of this neutral solution is added to a volumetric flask containing HCl:
- The mixture is poured into a 100 mL volumetric flask containing 2 moles of HCl.
- Since the initial solution was neutral (containing only NaCl and water), adding it does not change the HCl concentration already in the flask.
Calculate the final concentration of HCl in the volumetric flask:
- Total volume in the flask is 100 mL or 0.1 L.
- Moles of HCl initially in the flask = 2 mol.
- Concentration of HCl in the flask = \(\frac{\text{Number of moles}}{\text{Volume in L}} = \frac{2 \, \text{mol}}{0.1 \, \text{L}} = 20 \, \text{M}\).
From this calculation, we see that the solution in the flask becomes a 20 M HCl solution.
Thus, the correct answer is: 20 M HCl solution.
Approach Solution -2
First, let's analyze the reaction between NaOH and HCl in the beaker: Moles of NaOH = Molarity × Volume (in L) = 2 M × (10 / 1000) L = 0.02 moles Moles of HCl = Molarity × Volume (in L) = 1 M × (20 / 1000) L = 0.02 moles The reaction is: \[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \] From the stoichiometry, 1 mole of NaOH reacts with 1 mole of HCl.
Since we have 0.02 moles of each, they will completely neutralize each other.
The resulting solution in the beaker will contain NaCl and water, and will be neutral.
Now, 10 mL of this neutral solution (containing NaCl and water) is poured into a 100 mL volumetric flask containing 2 moles of HCl. The volume is then made up to 100 mL with distilled water.
The amount of HCl already present in the flask is 2 moles. The addition of 10 mL of the neutral solution from the beaker does not add any significant amount of HCl.
The total volume of the solution in the flask is 100 mL = 0.1 L.
The molarity of HCl in the flask is: \[ \text{Molarity of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of solution in L}} = \frac{2 \, \text{moles}}{0.1 \, \text{L}} = 20 \, \text{M} \]
The solution in the flask is 20 M HCl solution.
Top Questions on Solutions
- Concentrated nitric acid is labelled as 75%by mass. The volume in mL of the solution which contains 30 g of nitric acid is: Given: Density of nitric acid solution is 1.25 g/mL.
- 0.1 M solution of KI reacts with excess of \( \text{H}_2\text{SO}_4 \) and KIO\(_3\), according to the equation: \[ 5I^- + 6H^+ \rightarrow 3I_2 + 3H_2O \]Identify the correct statements:
- (A) 200 mL KI solution reacts with 0.004 mol of KIO3
- (B) 200 mL KI solution reacts with 0.006 mol of H2SO4
- (3) 0.5 L KI solution produced 0.005 mol of I2
- (4) Equivalent weight of KIO3 is equal to Molecular weight / 5
- Which of the following binary mixture does not show the behavior of minimum boiling azeotropes?
- 2 moles each of ethylene glycol and glucose are dissolved in 500 g of water. The boiling point of the resulting solution is: (Given: Ebullioscopic constant of water = $ 0.52 \, \text{K kg mol}^{-1} $)
Which of the following properties will change when system containing solution 1 will become solution 2 ?
Questions Asked in JEE Main exam
A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \( B \) exists into the page. The bar starts to move from the vertex at time \( t = 0 \) with a constant velocity. If the induced EMF is \( E \propto t^n \), then the value of \( n \) is _____.
- JEE Main - 2025
- Magnetic Field
- A physical quantity \( Q \) is related to four observables \( a \), \( b \), \( c \), and \( d \) as follows: \[ Q = \frac{a b^4}{c d^2} \] Where:
- \( a = (60 \pm 3) \, \text{Pa} \),
- \( b = (20 \pm 0.1) \, \text{m} \),
- \( c = (40 \pm 0.2) \, \text{N·s/m}^2 \),
- \( d = (50 \pm 0.1) \, \text{m} \).
Then the percentage error in \( Q \) is:- JEE Main - 2025
- Error analysis
- Let circle $C$ be the image of
$$ x^2 + y^2 - 2x + 4y - 4 = 0 $$
in the line
$$ 2x - 3y + 5 = 0 $$
and $A$ be the point on $C$ such that $OA$ is parallel to the x-axis and $A$ lies on the right-hand side of the centre $O$ of $C$.
If $B(\alpha, \beta)$, with $\beta < 4$, lies on $C$ such that the length of the arc $AB$ is $\frac{1}{6}$ of the perimeter of $C$, then $\beta - \sqrt{3}\alpha$ is equal to: - Assume a living cell with 0.9%(\(w/w\)) of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data up to first decimal place only) The cell will:
- JEE Main - 2025
- osmosis
Given below are two statements:
Statement (I):
are isomeric compounds.
Statement (II):
are functional group isomers.
In the light of the above statements, choose the correct answer from the options given below:- JEE Main - 2025
- Isomerism