0.01 mole of an organic compound (X) containing 10% hydrogen, on complete combustion produced 0.9 g . Molar mass of (X) is ______ g . [2025]

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Solution

Q.
0.01 mole of an organic compound (X) containing 10% hydrogen, on complete combustion produced 0.9 g $H_{2} O$ . Molar mass of (X) is ______ g ${mol}^{- 1}$ . [2025]

Ans.
(100)

Moles of water = $\frac{Given mass of H_{2} O}{Molar mass of H_{2} O} = \frac{0.9}{18} mol = 0.05 mol$

Moles of H $= 2 \times moles of water = 2 \times 0.05 mol = 0.1 mol$

Mass of H = $Moles of H \times molar mass of H = 0.1×1 g=0.1 g$

So 0.1 g H is present in 0.01 mol of an organic compound. Thus 10 g of H is present in 1 mol of the compound.

$% of H = \frac{Mass of H in 1 mol compound}{Molar mass of compound} \times 100$

$10 = \frac{10}{Molar mass of compound} \times 100$

Molar mass of compound = 100 g ${mol}^{- 1}$

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