Question

13.5 mL of a solution, made by mixing 6 g of a monobasic acid (mol. mass = M) in 1 L H₂0, was completely neutralised by 10 mL of a decinormal Ca(OH)₂ solution. Then VM is____

Answer 1

To determine VM, the molar mass of the monobasic acid and the volume of its solution that was titrated are needed. The calculation involves finding the moles of
`Ca(OH)_2`and hence the moles of the monobasic acid, followed by the molar mass and then VM. The calculated value of VM is 162g·L/mol.

Step-by-step explanation:

The student's question involves a titration calculation to find VM, which stands for the product of the molar mass (M) of a monobasic acid and the volume (V) of the acid solution that was titrated. We are given that 13.5 mL of the monobasic acid solution was neutralized by 10 mL of a decinormal (0.1 N)
`Ca(OH)_2` solution.

To find the molar mass, we use the formula:

The volume of the acid solution in liters is 0.0135L. The normality of
`Ca(OH)_2` can be used to find moles because it is equivalent to molarity for a monobasic acid. Since
`Ca(OH)_2` is dibasic, its molarity is half its normality, which is 0.05M. Thus, moles of
`Ca(OH)_2`= 0.01L × 0.05M = 0.0005 moles. This is also the moles of the acid. So, M = 6g / 0.0005 moles = 12000g/mol. Therefore, VM = 13.5mL × 12000g/mol = 162g·L/mol.