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.
We are given that the concentration of the nitric acid solution is 75% by mass. This means that for every 100 g of solution, there are 75 g of nitric acid. We are asked to find the volume of the solution that contains 30 g of nitric acid. First, calculate the total mass of the solution required to get 30 g of nitric acid: \[ {Mass of solution} = \frac{30 \, {g}}{0.75} = 40 \, {g} \] Now, using the density of the solution, which is 1.25 g/mL, calculate the volume of the solution: \[ {Volume} = \frac{40 \, {g}}{1.25 \, {g/mL}} = 32 \, {mL} \] Thus, the volume required is 40 mL.
Arrange the following in increasing order of solubility product:
\[ {Ca(OH)}_2, {AgBr}, {PbS}, {HgS} \]
Match List - I with List - II.
List - I (Saccharides) List - II (Glycosidic linkages found)
(A) Sucrose (I) \( \alpha 1 - 4 \)
(B) Maltose (II) \( \alpha 1 - 4 \) and \( \alpha 1 - 6 \)
(C) Lactose (III) \( \alpha 1 - \beta 2 \)
(D) Amylopectin (IV) \( \beta 1 - 4 \)
Choose the correct answer from the options given below:
Match List - I with List - II.
List - I (Complex) | List - II (Hybridisation) |
---|---|
(A) \([\text{CoF}_6]^{3-}\) | (I) \( d^2 sp^3 \) |
(B) \([\text{NiCl}_4]^{2-}\) | (II) \( sp^3 \) |
(C) \([\text{Co(NH}_3)_6]^{3+}\) | (III) \( sp^3 d^2 \) |
(D) \([\text{Ni(CN}_4]^{2-}\) | (IV) \( dsp^2 \) |
Choose the correct answer from the options given below:
The magnitude of heat exchanged by a system for the given cyclic process ABC (as shown in the figure) is (in SI units):