1 M and 2.5 litre NaOH solution mixed with another 0.5 M and 3 litre NaOH solution. Then find out molarity of resultant solution
The correct option is (C) : 0.73 M
Mole of 2.5 L of 1M NaOH =2.5×1=2.5
Mole of 3.0 L of 0.5 M NaOH =3.0×0.5=1.5
Total moles of NaOH in solution =2.5+1.5=4.0 (Total volume of solution =2.5+3.0=5.5L)
Thus, M1 ×V1=M2 ×V2 4.0=M2 ×5.5 Molarity of resultant solution = M2 = \(\frac{4.0}{5.5}\) =0.73M
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
What is Microalbuminuria ?
The output (Y) of the given logic implementation is similar to the output of an/a …………. gate.
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: