$5 \,g$ of $NaOH$ was dissolved in deionized water to prepare a $450 \,mL$ stock solution What volume (in $mL$ ) of this solution would be required to prepare $500 \,mL$ of $01\, M$ solution? Given: Molar Mass of $Na, O$ and $H$ is 23,16 and $1 \,g\, mol ^{-1}$ respectively
Correct Answer: 180
Approach Solution - 1
The correct answer is 180.
To prepare 500 mL of 0.1 M NaOH solution, we need to calculate the volume of the stock solution required.
First, let's calculate the number of moles of NaOH in 500 mL of 0.1 M solution:
Number of moles of NaOH = Molarity × Volume (in L) Number of moles of NaOH = 0.1 × 0.5 Number of moles of NaOH = 0.05
Next, let's calculate the number of moles of NaOH in the stock solution:
Number of moles of NaOH in stock solution = (mass of NaOH / molar mass of NaOH) = (5 g / 40 g/mol) = 0.125 mol
Now, we can use the following formula to calculate the volume of the stock solution required:
Volume of stock solution = (Number of moles required / Number of moles in stock solution) × Volume of stock solution
Plugging in the values, we get:
Volume of stock solution = (0.05 / 0.125) × 0.45 L Volume of stock solution = 0.18 L = 180 mL
Therefore, 180 mL of the NaOH stock solution is required to prepare 500 mL of 0.1 M NaOH solution.
Approach Solution -2
The correct answer is 180
\(M=\frac{5}{40} \times \frac{1000}{450}\)
\(M_1V_1=M_2V_2\)
\((\frac{5}{40} \times \frac{1000}{450})\times V_1=0.1 \times 500\)
\(V_1=180\)
\(\therefore\) volume required is 180
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Solutions Questions
- What weight of glucose must be dissolved in 100 g of water to lower the vapour pressure by 0.20 mm Hg?
(Assume dilute solution is being formed)
Given : Vapour pressure of pure water is 54.2 mm Hg at room temperature. Molar mass of glucose is 180 g mol–1 . - Solubility of \(Ca_3(PO_4)_2\) in \(100\) \(mL\) of pure water is \(W\) gm. Find out \(K_{sp}\) of \(Ca_3(PO_4)_2\) is: (M: Molecular mass of \(Ca_3(PO_4)_2\))
- We are given with three NaCl samples and their Van't Hoff factors. Which is correct option ?
Sample Van't Haff Factor Sample - 1 (0.1 M) \(i_1\) Sample - 2 (0.01 M) \(i_2\) Sample - 3 (0.001 M) \(i_2\) - Ethylene glycol of \(x\) \(kg\) is mixed with \(18.6\) \(kg\) of solvent, \(24 °C\) depression in freezing point takes place. Calculate value of \(x\). (Given: \(K_1\) = \(1.6 °C/molal\) M.W. of ethylene glycol = \(62\) \(g/mol\))
- $\text{Volume of } 3 \, \text{M NaOH (formula weight 40 g mol}^{-1}\text{) which can be prepared from } 84 \, \text{g of NaOH is } \_\_\_\_\_ \times 10^{-1} \, \text{dm}^3.$
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
Concepts Used:
Solutions
A solution is a homogeneous mixture of two or more components in which the particle size is smaller than 1 nm.
For example, salt and sugar is a good illustration of a solution. A solution can be categorized into several components.
Types of Solutions:
The solutions can be classified into three types:
- Solid Solutions - In these solutions, the solvent is in a Solid-state.
- Liquid Solutions- In these solutions, the solvent is in a Liquid state.
- Gaseous Solutions - In these solutions, the solvent is in a Gaseous state.
On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types:
- Unsaturated Solution- A solution in which more solute can be dissolved without raising the temperature of the solution is known as an unsaturated solution.
- Saturated Solution- A solution in which no solute can be dissolved after reaching a certain amount of temperature is known as an unsaturated saturated solution.
- Supersaturated Solution- A solution that contains more solute than the maximum amount at a certain temperature is known as a supersaturated solution.