What volume in milliliters of 100 proof vodka (50% v/v ethanol) would be likely to kill a typical 360-g rat?

The given question is incomplete, the complete question is:

The LD50 value for ethyl alcohol given orally to rats is 10.3 g/kg. What volume in milliliters of 100 proof vodka (50% v/v ethanol) would be likely to kill a typical 360-g rat?

The correct answer is 9.4 ml.

Explanation:

Based on the given information, in rats the LD50 of ethyl alcohol is 10.3 g/kg.

As 1 Kg is 1000 grams

Now the value of LD50 of ethyl alcohol in rats is 10.3 grams/ 1000 grams

The volume of the lethal dose of vodka for a 360 gram rat would be,  

= 360 g × (10.3 g/1000 g)

= 3708 × 10⁻³ g

= 3.708 g

Now the volume of vodka can be determined by using the formula,  

V = mass/density

The density of ethanol is 0.789 g/ml

The volume would be,  

V = 3.708 g/0.789 g/ml

V = 4.70 ml

As 50 percent of Vodka is ethanol, the volume of vodka needed to kill a rat is,  

= 2 × Volume of ethanol

= 2 × 4.70 g/ml

= 9.4 ml

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

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