Vanillin (used to flavor vanilla ice cream and other foods) is the substance whose aroma the human nose detects in the smallest amount. The threshold limit is 2.0 x 10^-11 g per liter of air. If the current price of 50 g of vanillin is $112, determine the cost to supply enough vanillin so that the aroma could be detected in a large aircraft hangar with a volume of 5.0 x 10^7 ft^3.

Threshold limit of vanillin in air is [tex]2.0\times 10^{-11}g[/tex] per litre means there should be [tex]2.0\times 10^{-11}g[/tex] of vanillin in 1L of air to detect aroma of vanillin.

[tex]1ft^{3}=28.32L[/tex]

So, [tex]5.0\times 10^{7}ft^{3}=(5.0\times 10^{7}\times 28.32)L[/tex]

So amount of vanillin should be present to detect = [tex](2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32)g[/tex]

As cost of 50 g vanillin is [tex]112\$[/tex] therefore cost of [tex](2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32)g[/tex]vanillin = [tex](2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32\times 112)\$ = 3.2\$[/tex]

adioabiola adioabiola · Answer 2 · 2021-11-15T13:18:33+01:00

The cost to supply enough Vanillin so that the aroma could be detected in a large aircraft hangar with a volume of 5.0 x 10^7 ft^3 is; $0.063

Conversion;

1 ft³ = 28.32 L.

Therefore, 5.0 x 10^7 ft^3. = 1.416 × 10⁹ L.

However, since the threshold limit is 2.0 x 10^-11 g per liter of air;

The quantity of Vanillin used in 1.416 × 10⁹ L of air is;

= 1.416 × 10⁹ L × 2.0 x 10^-11 g/L

= 2.832 × 10-² g of Vanillin.

Finally, Since 50 g of Vanillin costs $112;

Therefore, 2.832 × 10-² g of Vanillin will cost;

= (2.832 × 10-² g × 112)/50g = $0.063