We are given that the molar mass of the hydrocarbon (X) is 80 g/mol and it contains 90% carbon by mass. To find the degree of unsaturation, we first need to determine the empirical formula of the compound.
The mass of carbon in 80 g of the compound is 90% of 80, which is:
Mass of carbon = \( 0.90 \times 80 = 72 \, \text{g} \)
The mass of hydrogen is the remainder, which is:
Mass of hydrogen = \( 80 - 72 = 8 \, \text{g} \)
Now, calculate the moles of carbon and hydrogen using their respective atomic masses (C = 12 g/mol and H = 1 g/mol):
Moles of carbon = \( \frac{72 \, \text{g}}{12 \, \text{g/mol}} = 6 \, \text{mol} \)
Moles of hydrogen = \( \frac{8 \, \text{g}}{1 \, \text{g/mol}} = 8 \, \text{mol} \)
The mole ratio of carbon to hydrogen is 6:8, which simplifies to 3:4. Therefore, the empirical formula of the compound is C3H4.
The molar mass of the empirical formula C3H4 is:
Molar mass of C3H4 = \( 3 \times 12 + 4 \times 1 = 36 + 4 = 40 \, \text{g/mol} \)
The given molar mass of the compound is 80 g/mol. To find the molecular formula, divide the molar mass of the compound by the molar mass of the empirical formula:
Ratio = \( \frac{80}{40} = 2 \)
This means the molecular formula is twice the empirical formula, or C6H8.
The degree of unsaturation (DU) can be calculated using the formula:
\[ \text{DU} = \frac{2C + 2 - H}{2} \]
For C6H8, we have:
DU = \( \frac{2(6) + 2 - 8}{2} = \frac{12 + 2 - 8}{2} = \frac{6}{2} = 3 \)
The hydrocarbon (X) with a molar mass of 80 g/mol and 90% carbon has a degree of unsaturation of 3.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are


What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,