Chemistry, 23.03.2020 21:31 msalecora

A sample of a monoprotic acid (HA) weighing 0.384 g is dissolved in water and the solution is titrated with aqueous NaOH. If 30.0 mL of 0.100 M NaOH is required to reach the equivalence point, what is the molar mass of HA?

(a) 211 g/mol
(b) 128 g/mol
(c) 81.0 g/mol
(d) 37.0 g/mol
(e) 20.3 g/mol

Answers: 2

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Answer from: tangia

The correct answer is option b.

Explanation:

Moles of NaOH = n

Volume of NaOH solution = 30.0 mL =0.030 L ( 1mL =0.001 L)

Molarity of the NaOH solution = 0.100 M

$Molarity=\frac{Moles}{Volume(L)}$

$n=0.100 M\times 0.030 L=0.00300 mol$

$HA(aq)+NaOH(aq)\rightarrow NaA(aq)+H_2O(aq)$

According to reaction , 1 mol of NaOH reacts with 1 mole of HA, then, 0.00300 moles of NaOH will react with :

$\frac{1}{1}\times 0.00300 mol=0.00300 mol$ of HA

Moles of HA = 0.00300 mol

Mass of HA = 0.384 g

Molar mass of HA = M

Mass of 0.00300 moles of HA = 0.384 g

$0.00300 mol\times M=0.384 g$

$M=\frac{0.384 g}{0.00300 mol}=128 g/mol$

128 g/mol is the molar mass of HA.

Answer from: Quest

Answer: abeexplanation:

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.

Answer from: Quest

I'm not sure but i will try to find out

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A sample of a monoprotic acid (HA) weighing 0.384 g is dissolved in water and the solution is titrat...