-
We need your support!
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
chem p5 help needed!urgent!
- Thread starter nisurju
- Start date
I am not very gud at explaining but i'll try my best
Part d
We use 30 cm3 of 2.00 moldm-3 so we have 0.06 moles
c=n/v so n = 0.03*2 = 0.06
Hydrochloric acid, nitric acid and ethanoic acid are all monoprotic acids so number of moles for these acids will be 0.06
You can select a volume and concentration that give you 0.06 moles '
For example 20cm3 of 3.0 moldm-3
Sulfuric acid and ethanedioc acid are both diprotic acids so ratio of NaoH to either of them is 2:1 thereby number of moles of the acid will be 0.03
You have to keep in mind that total volume shoukd be equal so we can use 20cm3 of 1.5moldm-3
Part e
to solve this part we need to first figure out the mass pf solid that we will use
we are given that final volume is 250 cm3 and we used a concentration of 1.5 moldm-3 so
c=n/v n= 1.5*0.25 n= 0.375 moles
molar mass of the acid is 126 so n=m/Mr thereby m = 126* 0.375 n= 47.25g
Use a digital balance to weigh 47.25g of hydrated crystalline solid into a 250 cm3 beaker and add 100cm3 of water, stir with a glass rod till all the solid dissolves> transfer the solution to a volumetric flask. rinse the beaker and rod atleast twice with water and add this to the volumetric flask. fill the voulmetric flask with distill water upto the lower meniscus of the mark. Put the stopper, swirl and invert the flask several times to mix the solution.
Part d
We use 30 cm3 of 2.00 moldm-3 so we have 0.06 moles
c=n/v so n = 0.03*2 = 0.06
Hydrochloric acid, nitric acid and ethanoic acid are all monoprotic acids so number of moles for these acids will be 0.06
You can select a volume and concentration that give you 0.06 moles '
For example 20cm3 of 3.0 moldm-3
Sulfuric acid and ethanedioc acid are both diprotic acids so ratio of NaoH to either of them is 2:1 thereby number of moles of the acid will be 0.03
You have to keep in mind that total volume shoukd be equal so we can use 20cm3 of 1.5moldm-3
Part e
to solve this part we need to first figure out the mass pf solid that we will use
we are given that final volume is 250 cm3 and we used a concentration of 1.5 moldm-3 so
c=n/v n= 1.5*0.25 n= 0.375 moles
molar mass of the acid is 126 so n=m/Mr thereby m = 126* 0.375 n= 47.25g
Use a digital balance to weigh 47.25g of hydrated crystalline solid into a 250 cm3 beaker and add 100cm3 of water, stir with a glass rod till all the solid dissolves> transfer the solution to a volumetric flask. rinse the beaker and rod atleast twice with water and add this to the volumetric flask. fill the voulmetric flask with distill water upto the lower meniscus of the mark. Put the stopper, swirl and invert the flask several times to mix the solution.
- Messages
- 1,148
- Reaction score
- 166
- Points
- 73
Correction in above answer:
In part d) because NaOH need to be in excess, use concentrations and volumes of acids that give you less number of moles than 0.06 (moles of NaOH).
In part d) because NaOH need to be in excess, use concentrations and volumes of acids that give you less number of moles than 0.06 (moles of NaOH).