Given that 25ml of a solution contains 159g of sodium carbonate per dm3 was required for complete neutralization of 4.725g of monobasic acid at s.t.p. (Na=23, C=12, O=16, molar gas volume at s.t.p=22.4dm3) . find the relative molecular mass of the acid and the volume of carbon dioxide produced.
To find the relative molecular mass of the acid and the volume of carbon dioxide produced, we need to follow a series of steps:
Step 1: Calculate the number of moles of sodium carbonate used.
Given that 25 ml of the solution contains 159g of sodium carbonate per dm3, we can calculate the number of moles by using the formula:
moles of sodium carbonate = mass / molar mass
where,
mass = 159g (given)
molar mass of sodium carbonate (Na2CO3) = (2 * atomic mass of Na) + atomic mass of C + (3 * atomic mass of O)
= (2 * 23) + 12 + (3 * 16) = 106g/mol
So, moles of sodium carbonate = 159g / 106g/mol
Step 2: Calculate the number of moles of acid.
Given that 4.725g of monobasic acid were required for complete neutralization, we can calculate the number of moles of acid using the formula:
moles of acid = mass / molar mass
where,
mass = 4.725g (given)
molar mass of acid = moles of sodium carbonate / moles of acid
Step 3: Find the relative molecular mass of the acid.
To find the relative molecular mass of the acid, we need to determine the molar mass of the acid. Since we already know the number of moles of the acid calculated in the previous step, we can rearrange the formula to calculate the molar mass:
molar mass of acid = mass / moles of acid
Step 4: Calculate the volume of carbon dioxide produced.
To find the volume of carbon dioxide produced, we need to use the concept of molar gas volume at S.T.P (standard temperature and pressure). Given that the molar gas volume at S.T.P is 22.4 dm3/mol, we can use the formula:
volume of carbon dioxide = moles of carbon dioxide * molar gas volume at S.T.P
Now, let's plug in the values and calculate the relative molecular mass of the acid and the volume of carbon dioxide produced.