determine the volume, in ml, of a 381.9 - g sample of an alloy with density of 0.258 $\frac{lb}{in^{3}}$.\n0…

Explanation:

Step1: Recall density formula

The density formula is $\rho=\frac{m}{V}$, where $\rho$ is density, $m$ is mass and $V$ is volume. We can re - arrange it to $V = \frac{m}{\rho}$ for the first problem and $m=\rho V$ for the second problem. Also, note that $1\ cm^{3}=1\ mL$.

Step2: Solve for the volume of the alloy

Given $m = 381.9\ g$ and $\rho=0.258\frac{lb}{in^{3}}$. First, convert the density to $g/cm^{3}$. Since $1\ lb = 453.592\ g$ and $1\ in^{3}=16.3871\ cm^{3}$, then $\rho=0.258\frac{453.592\ g}{16.3871\ cm^{3}}\approx7.14\ g/cm^{3}$. Then $V=\frac{m}{\rho}=\frac{381.9\ g}{7.14\ g/cm^{3}}\approx53.5\ cm^{3}=53.5\ mL$.

Step3: Solve for the mass of glycerol

Given $\rho = 1.26\ g/cm^{3}$ and $V = 9.30\ mL=9.30\ cm^{3}$. Using $m=\rho V$, we have $m=1.26\ g/cm^{3}\times9.30\ cm^{3}=11.718\ g\approx11.7\ g$.

Answer:

7. 53.5mL
8. 11.7g