For question #6, using results from question #5. 5. Determine the masses (mg) of

Question

For question #6, using results from question #5.

5. Determine the masses (mg) of FD&C; Blue 1 and FD&C; Yellow 5 in a packet of lemon-lime Kool-Aid. A packet makes 2 quarts, or 1.89 L of Kool-Aid. FD&C; Yellow :C16Hg4NaOS2 [FD&C; Blue 1] in lemon-lime Kool-Aid [FD&C; Yellow 5] in lemon-lime Kool-Aid 3.50x101 mg Ok 4.68 mg Ok Your results are now shown above. 6. The LD50 for a substance is the armount of that substance that, given at once, causes the death of 50% ofthe test population (LD stands for "lethal dose"). For FD&C; Blue 1, the LDso is 300 mg/kg (rat); for FD&C; Yellow 5 the LDso is 12,750 mg/kg mouse Using your results from, Question #5, determine the minimum number of packets of Kool Aid an average adult (assume 70 kg) would have to consume to reach the LDso level for the dyes (assuming that the LDso levels for a rat and mouse correspond to the LDso level for a human). Report your answer to three significant figures (you may need to use scientific notation). packets Submt s velue is not correct. Please check your

Explanation / Answer

Let's find out first, the things given in the question:

LD50 value of FC&D blue 1 =300 mg/kg of rat

LD50 value of FC&D Yellow 5 =12,750 mg/kg of mouse

1 Packet of lemon-lime Kool-Aid contains:

FC&D blue 1= 4.68 mg

FC&D Yellow 5= 3.50 x 101 mg

It was given in the question that LD50 level of rat and mouse corresponds to the LD50 level of human and the weight of that human was assumed to be 70kg.

To get an equivalent dose of LD50 for both FC&D blue 1 and FC&D yellow as of rat and mouse respectively we should know the total dose corresponding to the weight of human i.e. 70 kg.

Therefore we can multiply the total weight with the corresponding LD50 values per kg.

FC&D blue 1 =300 x 70

            Or,      = 21000 mg/70 kg

FC&D Yellow 5 =12,750x 70

            Or,           = 892,500 mg/70 kg

So we get, 21000 mg/70 kg of FC&D blue 1 and 892,500 mg/70 kg FC&D Yellow 5

To determine the minimum number of Kool-Aid packet we can solve this:

For, FC&D blue 1= Total LD50 value/Total weight/ Mass ofFC&D blue 1 in 1 Kool-Aid Packet

=21000/4.68

= 4.487 x103 packets for FC&D blue 1

To determine the minimum number of Kool-Aid packet we can solve this:

For, FC&D Yellow 5 = Total LD50 value/Total weight/ Mass ofFC&D blue 1 in 1Kool                                  Aid Packet

=892,500 /3.50 x101

= 25.5 x103 packets for FC&D Yellow 5.

Hence, an average human adult with 70kg weight would consume 4.487 x103 packets of Kool-Aid for FC&D blue 1 and 25.5 x103 packets of Kool-Aid for FC&D Yellow 5 respectively to reach the LD50 level of the dyes.

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