Identification of the Monohydric Alcohol and Carbonyl Compound

Step 1: Calculation of the Molecular Weight of the Monohydric Alcohol
To identify the monohydric alcohol, we need to calculate its molecular weight using the given mass of 0.44g. Assuming the density of the alcohol is 1g/mL, we have:

Mass = Volume x Density
Volume = Mass / Density = 0.44g / 1g/mL = 0.44mL

Molecular weight = Mass / Volume = 0.44g / 0.44mL = 1g/mL

Therefore, the molecular weight of the monohydric alcohol is 1g/mol.

Step 2: Calculation of the Number of Moles of Methane Produced
When the monohydric alcohol is added to methyl magnesium iodide in ether, it liberates 112cm^3 of methane at STP. We can calculate the number of moles of methane produced using the ideal gas law:

PV = nRT
n = PV / RT

where P = pressure, V = volume, n = number of moles, R = gas constant, and T = temperature.

At STP, P = 1 atm, V = 112cm^3 = 0.112L, R = 0.08206 L atm/mol K, and T = 273 K.

n = (1 atm x 0.112L) / (0.08206 L atm/mol K x 273 K) = 0.0042 mol

Therefore, 0.0042 mol of methane is produced.

Step 3: Calculation of the Number of Moles of the Monohydric Alcohol
The balanced chemical equation for the reaction between the monohydric alcohol and methyl magnesium iodide is:

ROH + CH3MgI → RH + CH4 + MgI2

where ROH is the monohydric alcohol and RH is the corresponding alkane.

From the equation, we can see that 1 mole of the monohydric alcohol produces 1 mole of the corresponding alkane. Therefore, the number of moles of the monohydric alcohol is also 0.0042 mol.

Step 4: Identification of the Monohydric Alcohol
To identify the monohydric alcohol, we need to consider its molecular weight and the number of moles produced. From Step 1 and 3, we know that the molecular weight of the monohydric alcohol is 1g/mol and the number of moles produced is 0.0042 mol.

The monohydric alcohols with a molecular weight of 1g/mol are methanol (CH3OH) and ethynol (C2H5OH). However, the number of moles produced is very low, indicating that the alcohol is likely to be methanol, which is a primary alcohol.

Step 5: Formation of a Carbonyl Compound
The carbonyl compound formed from the monohydric alcohol reacts with Tollens' reagent (ammoniacal silver nitrate) to give a silver mirror. This indicates the presence of an aldehyde or a ketone. Since methanol is a primary alcohol