Question

a 1.84 mg sample of polyhydric alcoholic compound ' x ' of molar mass 92.0 g / mol gave 1.344 ml of H₂ gas at stp. the number of alcoholic hydrogen present in compound ' x ' is

Answer 1

The number of alcoholic hydrogen present in compound 'x' is 0.04 mmol.

Step-by-step explanation:

The number of alcoholic hydrogen present in compound 'x' can be determined using stoichiometry and the given information. First, calculate the moles of ethylene glycol using its molar mass. Given that the sample mass is 1.84 mg and the molar mass is 92.0 g/mol, the moles of ethylene glycol is:

moles = mass / molar mass

moles = 1.84 mg / 92.0 g/mol = 0.02 mmol

This means that there are 0.02 mmol of ethylene glycol in the sample. Since ethylene glycol has two alcoholic hydrogens per molecule, multiply the moles of ethylene glycol by 2 to calculate the moles of alcoholic hydrogen:

moles of alcoholic hydrogen = moles of ethylene glycol x 2

moles of alcoholic hydrogen = 0.02 mmol x 2 = 0.04 mmol

Therefore, the number of alcoholic hydrogen present in compound 'x' is 0.04 mmol.