Question #c233b

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Question #c233b

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Answer 1 · 2015-02-04T23:37:00

In order to solve this problem you must use a couple of unit conversions in the form of conversion factors. Likewise, you'll need the formula for density, which is defined as mass per unit of volume.

$ρ = \frac{m}{V} \Rightarrow V = \frac{m}{ρ}$ $rho = m/V => V = m/(rho)$

You can either convert pounds to kg or kg to pounds, but you must convert $m^{3}$ $"m"^(3)$ to $mL$ $"mL"$ for the expression of the sample's volume. So,

$1.36 \cdot 10^{4} \frac{kg}{m^{3}} \cdot \frac{2.2046 lbs}{1 kg} \cdot \frac{{1 m}^{3}}{10^{6} mL} = 3.00 \cdot 10^{- 2} \frac{lbs}{mL}$ $1.36 * 10^(4)"kg"/"m"^(3) * ("2.2046 lbs")/("1 kg") * ("1 m"^(3))/(10^(6)"mL") = 3.00 * 10^(-2)"lbs"/"mL"$

Now all you have to do is use the formula for density to determine volume

$V = \frac{m}{ρ} = \frac{1.00 lbs}{3.00 \cdot 10^{- 2} \frac{lbs}{mL}} = 33.3 mL$ $V = m/(rho) = ("1.00 lbs")/(3.00 * 10^(-2)"lbs"/"mL") = "33.3 mL"$

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Question #c233b

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