Calculate the Molarity of a Solution having density = \(1.5\) \(g/ml\). \(\%(w/w)\) of Solute is \(36\%\) and Molecular weight of Solute is \(36\) \(g/mol\)
Approach Solution - 1
Approach Solution -2
Assume the mass of the solution = 100g
\(\begin{aligned} & \text { Mass of solute }=36 \mathrm{gm} \\ & \text { Moles of solute }=1 \\ &Molarity= \frac{moles of solute}{volume of solution in liters}\\& \text { Molarity }=\frac{1 \times 1000}{\left(\frac{100}{1.5}\right)}=\frac{1000}{100} \times 1.5=15 \end{aligned}\)
So, the answer is 15.
Top Questions on Mole concept and Molar Masses
- When '$x \times 10^{-2} \, \text{mL}$' methanol (molar mass $= 32 \, \text{g}$; density $= 0.792 \, \text{g/cm}^3$) is added to $100 \, \text{mL}$ water (density $= 1 \, \text{g/cm}^3$), the following diagram is obtained.
x = __________ (nearest integer)
[Given: Molal freezing point depression constant of water at $273.15 \, \text{K}$ is $1.86 \, \text{K kg mol}^{-1}$]- JEE Main - 2024
- Chemistry
- Mole concept and Molar Masses
- A sample of \( \text{CaCO}_3 \) and \( \text{MgCO}_3 \) weighed 2.21 g is ignited to constant weight of 1.152 g. The composition of the mixture is:
(Given molar mass in g mol\(^{-1}\))- JEE Main - 2024
- Chemistry
- Mole concept and Molar Masses
- Ail organic compound has 42.1% carbon, 6.4% hydrogen and remainder is oxygen. If its molecular weight is 342, then its molecular formula is :
- JEE Main - 2024
- Chemistry
- Mole concept and Molar Masses
- Molality $(m)$ of $3 \, \text{M}$ aqueous solution of $\text{NaCl}$ is:
(Given: Density of solution $= 1.25 \, \text{g mL}^{-1}$, Molar mass in $\text{g mol}^{-1}$: $\text{Na} = 23, \, \text{Cl} = 35.5$)- JEE Main - 2024
- Chemistry
- Mole concept and Molar Masses
- Number of moles of \(H^+\) required by \(1\) mole \(MnO_4^-\) to oxidize oxalate ion to \(CO_2\) is ____ .
- JEE Main - 2024
- Chemistry
- Mole concept and Molar Masses
Questions Asked in JEE Main exam
- If the domain of the function \(f(x)\) = \(\frac{\sqrt{x^2-25}}{(4-x^2)}+ \log(x^2+2x-15)\) is \((-โ, ฮฑ) โช (ฮฒ, โ)\), then \(ฮฑ^2+ฮฒ^2\) is equal to \(b\)
- JEE Main - 2024
- Functions
- A group of 40 students appeared in an examination of 3 subjects โ Mathematics, Physics & Chemistry. It was found that all students passed in at least one of the subjects, 20 students passed in Mathematics, 25 students passed in Physics, 16 students passed in Chemistry, at most 11 students passed in both Mathematics and Physics, at most 15 students passed in both Physics and Chemistry, at most 15 students passed in both Mathematics and Chemistry. The maximum number of students passed in all the three subjects is ________.
- JEE Main - 2024
- Set Theory
- Consider the line \( L \) passing through the points \( (1, 2, 3) \) and \( (2, 3, 5) \). The distance of the point \[ \left( \frac{11}{3}, \frac{11}{3}, \frac{19}{3} \right) \] from the line \( L \) along the line \[ \frac{3x - 11}{2} = \frac{3y - 11}{1} = \frac{3z - 19}{2} \] is equal to:
- JEE Main - 2024
- 3D Geometry
- Primary coil of a transformer is connected to 220 V ac. Primary and secondary turns of the transforms are 100 and 10 respectively. Secondary coil of transformer is connected to two series resistance shown in shown in figure. The output voltage (V0) is :
- JEE Main - 2024
- Resistance
- A 2A current-carrying straight metal wire of resistance 1 ฮฉ, resistivity \( 2 \times 10^{-6} \, \Omega \, \text{m} \), area of cross-section \( 10 \, \text{mm}^2 \), and mass 500 g is suspended horizontally in mid-air by applying a uniform magnetic field \( \vec{B} \). The magnitude of \( B \) is ________ \( \times 10^{-1} \, \text{T} \) (given, \( g = 10 \, \text{m/s}^2 \)).
- JEE Main - 2024
- Magnetic Effects of Current and Magnetism
Concepts Used:
Mole Concept
In the field of chemistry, a mole is defined as the amount of a substance that contains exactly 6.02214076 * 1023 โelementary entitiesโ of the given substance.
The number 6.02214076*1023 is popularly known as the Avogadro constant and is often denoted by the symbol โNAโ. The elementary entities that can be represented in moles can be atoms, molecules, monoatomic/polyatomic ions, and other particles (such as electrons).
For example, one mole of a pure carbon-12 (12C) sample will have a mass of exactly 12 grams and will contain 6.02214076*1023 (NA) number of 12C atoms. The number of moles of a substance in a given pure sample can be represented by the following formula:
n = N/NA
Where n is the number of moles of the substance (or elementary entity), N is the total number of elementary entities in the sample, and NA is the Avogadro constant.
The word โmoleโ was introduced around the year 1896 by the German chemist Wilhelm Ostwald, who derived the term from the Latin word moles meaning a โheapโ or โpile.
The mole concept refers to the atomic mass of a mole that is measured in grams. The gram atomic mass of an element is known as a mole. The mole concept combines the mass of a single atom or molecule in a.m.u. to the mass of a large group of comparable molecules in grams. Atomic mass is the mass of a single atom, whereas molecular mass is the mass of a group of atoms.
The Formula of Mole Concept
The number of units that make up a mole has been calculated to be 6.022 ร10ยฒยณ. The fundamental constant is also known as Avogadro's number (NA) or Avogadro constant. This constant is appropriately represented in chemistry using an explicit unit termed per mole.
Number of Moles = (Mass of the Sample)/(Molar Mass)
Read More: Mole Fraction