Question:
The molar mass of the water insoluble product formed from the fusion of chromite ore \(FeCr_2\text{O}_4\) with \(Na_2\text{CO}_3\) in presence of \(O_2\) is ....... g mol\(^{-1}\):
The molar mass of the water insoluble product formed from the fusion of chromite ore \(FeCr_2\text{O}_4\) with \(Na_2\text{CO}_3\) in presence of \(O_2\) is ....... g mol\(^{-1}\):
Show Hint
To calculate the molar mass of a compound, add the molar masses of the constituent atoms.
Updated On: Jan 14, 2026
Hide Solution
Verified By Collegedunia
Correct Answer: 196
Solution and Explanation
- The fusion of chromite ore (\( \text{FeCr}_2\text{O}_4 \)) with Na₂CO₃ and oxygen leads to the formation of Fe₂O₃ and Cr₂O₃ as products.
- The molar mass of the water-insoluble product (Cr₂O₃) is calculated: \[ \text{Molar mass of Cr}_2\text{O}_3 = 2(52) + 3(16) = 196 \, \text{g/mol} \] Thus, the molar mass of the water-insoluble product is 196 g/mol.
- The molar mass of the water-insoluble product (Cr₂O₃) is calculated: \[ \text{Molar mass of Cr}_2\text{O}_3 = 2(52) + 3(16) = 196 \, \text{g/mol} \] Thus, the molar mass of the water-insoluble product is 196 g/mol.
Was this answer helpful?
1
1
Top Questions on Mole concept and Molar Masses
- 80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is:
- JEE Main - 2026
- Chemistry
- Mole concept and Molar Masses
- The following solutions were prepared by dissolving 1 g of solute in 1 L of the solution. Arrange the following solutions in decreasing order of their molarity: (A) Glucose (molar mass = 180 g mol$^{-1}$)
(B) NaOH (molar mass = 40 g mol$^{-1}$)
(C) NaCl (molar mass = 58.5 g mol$^{-1}$)
(D) KCl (molar mass = 7(4)5 g mol$^{-1}$)
- CUET (UG) - 2025
- Chemistry
- Mole concept and Molar Masses
- Calculate the mass in kg of 4.48 dm$^3$ carbon dioxide at STP.
- MHT CET - 2025
- Chemistry
- Mole concept and Molar Masses
- The carbohydrate used as a storage molecule in plants is
- CUET (UG) - 2025
- Chemistry
- Mole concept and Molar Masses
- Match List-I with List-II \[ \begin{array}{|l|l|l|} \hline \textbf{List-I} & \textbf{List-II} & \textbf{Matches} \\ \hline (A) \; \text{Cell constant} & (I) \; \text{cm}^{-1} & (A) - (I) \\ \hline (B) \; \text{Molar conductance} & (II) \; \Omega^{-1}\,\text{cm}^2\,\text{mol}^{-1} & (B) - (II) \\ \hline (C) \; \text{Specific conductance} & (III) \; \Omega^{-1}\,\text{cm}^{-1} & (C) - (III) \\ \hline (D) \; \text{Conductance} & (IV) \; \Omega^{-1} & (D) - (IV) \\ \hline \end{array} \]
- CUET (UG) - 2025
- Chemistry
- Mole concept and Molar Masses
View More Questions
Questions Asked in JEE Main exam
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$)
- JEE Main - 2026
- Rotational Mechanics
- If the coefficient of \( x \) in the expansion of \[ (ax^2 + bx + c)(1 - 2x)^{26} \] is \(-56\) and the coefficients of \( x^2 \) and \( x^3 \) are both zero, then \( a + b + c \) is equal to
- JEE Main - 2026
- Binomial theorem
The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\).
- JEE Main - 2026
- Current electricity
Method used for separation of mixture of products (B and C) obtained in the following reaction is:
- JEE Main - 2026
- p -Block Elements
- Consider light travelling from a medium A to medium B separated by a plane interface. If the light undergoes total internal reflection during its travel from medium A to B and the speed of light in media A and B are \(2.4 \times 10^8\) m/s and \(2.7 \times 10^8\) m/s, respectively, then the value of critical angle is :
- JEE Main - 2026
- Newtons Laws of Motion
View More Questions