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# moles! watch

3.88g of monoprotic acid was dissolved in water and the solution made up to 250cm3. 25cm3 of this solution was titrated with 0.095mol dm-3 NaOH solution, requiring 46.5cm3. Calculate the relative molecular mass of the acid.

my attempt was:
moles = (25/1000) x 0.095 mol dm3 = 2.38x10-3 moles
molar mass = 3.88g/ 2.38x10-3 moles = 1634 Mr
2. (Original post by holderbolder)
3.88g of monoprotic acid was dissolved in water and the solution made up to 250cm3. 25cm3 of this solution was titrated with 0.095mol dm-3 NaOH solution, requiring 46.5cm3. Calculate the relative molecular mass of the acid.

my attempt was:
moles = (25/1000) x 0.095 mol dm3 = 2.38x10-3 moles
molar mass = 3.88g/ 2.38x10-3 moles = 1634 Mr

Take it a step at a time.
-----------------------

Moles of NaOH = molarity x volume (litres)

Moles of acid = mol NaOH

... but this is in a 25 ml aliquot (sample of liquid)

So, total moles of acid in 250 cm3 = mol NaOH x 10

Relative mass = mass/moles = 3.88/(mol NaOH x 10)
------------------------------
try again
3. Being a noob at this, lemme give it a try to see how good his instructions were

Moles of NaOH = 0.095 mol dm-3 x (46.5/1000)

= 0.0044175 mol

Moles of HCL aliquot = Moles of NaOH

however

Moles of HCL aliquot (25cm3) x10 = Moles of HCL (250cm3)

= 0.044175

Relative molecular mass = 3.88g/0.044175mol

= 87.8 (3sf)

Rounded up to nearest integer = 88

Hope I got that right
4. (Original post by kiLLthriLL)
Being a noob at this, lemme give it a try to see how good his instructions were

Moles of NaOH = 0.095 mol dm-3 x (46.5/1000)

= 0.0044175 mol

Moles of HCL aliquot = Moles of NaOH

however

Moles of HCL aliquot (25cm3) x10 = Moles of HCL (250cm3)

= 0.044175

Relative molecular mass = 3.88g/0.044175mol

= 87.8 (3sf)

Rounded up to nearest integer = 88

Hope I got that right
most certainly did

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