$32\ g$ of oxygen and $3\ g$ of hydrogen are mixed and kept in a vessel at $760\ mm$ pressure and $0^{\circ}C$. The total volume occupied by the mixture will be nearly:

The number of moles of oxygen is 32 g / 32 g/mol = 1 mole, and the number of moles of hydrogen is 3 g / 2 g/mol = 1.5 moles. The total number of moles in the mixture is 1 + 1.5 = 2.5 moles. At STP, one mole of any ideal gas occupies 22.4 litres, so the total volume occupied by the mixture is 2.5 moles * 22.4 litres/mole = 56 litres. The total volume occupied by the mixture will be nearly 56 litre.

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