I\'ve solved part one a) mass = 0.5973 gm molar mass = 204.224 gm number of mole

Question

I've solved part one

a)
mass = 0.5973 gm
molar mass = 204.224 gm
number of moles = mass/molar mass
                            = 0.5973 / 204.224
                            = 2.925*10^-3 mol
b)
This is added in 250mL = 0.25 L water.
So this 250 mL will have 2.925*10^-3 mol

c)
When we take 10 mL of above solution,
number of moles that is getting transferred = 10*2.925*10^-3 / 250
                     =1.17*10^-4 mol
This many moles are there in 100 ml= 0.1 L

Molarity = number of moles/ volume
                = (1.17*10^-4) / (0.1)
                 =1.17*10^-3 M

I just want to double check my answer to part II. Please show work. Thanks!!!

(4.5pts) A student prepared a standard solution of potassium acid phthalate ('KHP. 204.224 ± 0.003 g/mol) according to the following lab procedure: (a) Mass of KHP was determined = 0.5973 g ( 0.0001 g) (weighing vessel tared to 0.0000 g omit this tare step from error propagation, however) (b) KHP was transferred to a 250.00 mL ( 0.12 mL) volumetric flask, dissolved and diluted to the (c) A 10.00 mL (± 0.02) aliquot of the solution from (b) was transferred to a 100.00 mL ( 007) (d) Solution diluted to mark of 100.00 mL volumetric flask and mixed we mark. volumetric using a 1o.oo mL volumetric pipet G Calculate the KHP solution molarity to four significant figures. Please show work or method of calculation. (ii) Using the propagation of errors technique, determine the uncertainty in the resulting concentration also to four significant figures (here they may not all be significant, but keep four anyway please.)

Explanation / Answer

a)
mass of KHP = 0.5973 gm
molar mass of KHP = 204.224 gm
number of moles = mass/molar mass
                            = 0.5973 / 204.224
                            = 2.926*10^-3 mol
b)
This is added in 250mL = 0.25 L water.
So this 250 mL will have 2.926*10^-3 mol

So molarity = 2.926*10^-3 mol / 0.25 = 0.0117 molar

c) When we are taking10 mL from above solution and make it upto 100mL the new molarity will be

M1V1 = M2V2

0.0117 X 10 = M2 X 100

M2 = 0.001177 molar

You are right!
When we take 10 mL of above solution,
number of moles that is getting transferred = 10*2.925*10^-3 / 250
                     =1.17*10^-4 mol
This many moles are there in 100 ml= 0.1 L

Molarity = number of moles/ volume
                = (1.17*10^-4) / (0.1)
                 =1.17*10^-3 M

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