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.
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
Observe the following data given in the table. (\(K_H\) = Henry's law constant)
| Gas | CO₂ | Ar | HCHO | CH₄ |
|---|---|---|---|---|
| \(K_H\) (k bar at 298 K) | 1.67 | 40.3 | \(1.83 \times 10^{-5}\) | 0.413 |
The correct order of their solubility in water is
Match List I with List II:
Choose the correct answer from the options given below:
In the circuit shown, assuming the threshold voltage of the diode is negligibly small, then the voltage \( V_{AB} \) is correctly represented by:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
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.
The solutions can be classified into three types:
On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types: