Question

problem: the density of a pure substance is its mass per unit volume. the density of cresol has been measured to be 1024 g/l. calculate the mass of 305. ml of cresol. be sure your answer has a unit symbol and the correct number of significant digits. solution: p = m / v write down the definition of density. m = p / v solve for mass. put in the data given: m = 305. / 1024 g use the calculator: m = 297.85....g round to the correct number of significant digits: m = 298 g no mistakes

Explanation:

Step1: Recall density formula

The formula for density is $
ho = \frac{m}{V}$, so to find mass $m$, we rearrange it to $m=
ho\times V$.

Step2: Convert volume units

The volume $V = 305\space mL$. Since $1\space L = 1000\space mL$, we convert $305\space mL$ to liters: $V=\frac{305}{1000}\space L = 0.305\space L$.

Step3: Substitute values into mass formula

Given $
ho = 1024\space \frac{g}{L}$ and $V = 0.305\space L$, we calculate $m$ as $m=
ho\times V=1024\space \frac{g}{L}\times0.305\space L$.

Step4: Perform the multiplication

$1024\times0.305 = 1024\times(0.3 + 0.005)=1024\times0.3+1024\times0.005 = 307.2+5.12 = 312.32\space g$. Wait, but let's check the original solution's mistake. Wait, the original solution had $m=\frac{305}{1024}\space g$ which is wrong. The correct formula is $m =
ho V$, so $
ho = 1024\space g/L$, $V = 305\space mL = 0.305\space L$, so $m = 1024\times0.305 = 312.32\space g$. But maybe the significant figures: density is 1024 (4 sig figs), volume 305 (3 sig figs), so the answer should have 3 sig figs. $312.32\approx312\space g$? Wait, no, 1024 is 4 sig figs, 305 is 3, so the result should have 3. Wait, maybe the volume was 305 mL, and density is 1024 g/L. Let's recalculate: $1024\space g/L\times0.305\space L = 1024\times0.305$. Let's do 1024*0.3 = 307.2, 1024*0.005=5.12, total 312.32. Rounding to 3 sig figs: 312 g? Wait, but the original solution's first step had $p=\frac{m}{V}$ (which is correct for density definition), then solving for mass should be $m = pV$, not $m=\frac{V}{p}$. So the first mistake is in the formula for $m$: it should be $m = p\times V$, not $m=\frac{V}{p}$. Then, substituting $p = 1024\space g/L$ and $V = 305\space mL = 0.305\space L$, so $m = 1024\times0.305 = 312.32\space g$, which rounds to 312 g (3 sig figs) or maybe 312. g? Wait, maybe the volume was 305 mL, and 1 L = 1000 mL, so $V = 305\space mL = 0.305\space L$. So correct $m = 1024\times0.305 = 312.32\space g$, round to 3 sig figs: 312 g. But the original solution had $m=\frac{305}{1024}\space g$ which is incorrect. So the correct steps:

1. Density formula: $

ho=\frac{m}{V}\implies m =
ho V$.

1. Convert $V = 305\space mL$ to $L$: $V=\frac{305}{1000}\space L = 0.305\space L$.
2. Substitute $

ho = 1024\space g/L$ and $V = 0.305\space L$: $m = 1024\times0.305 = 312.32\space g$.

1. Round to 3 significant figures (since 305 has 3, 1024 has 4, so result has 3): $m\approx312\space g$.

But the original solution's first step for $m$ was wrong (used $m=\frac{V}{p}$ instead of $m = pV$). So correcting that, the mass should be calculated as $m =
ho V$.

Answer:

The first mistake is in the formula for mass: it should be $m =
ho V$ (or $m = pV$ using the problem's notation) instead of $m=\frac{V}{p}$. The correct mass calculation is $m = 1024\space \frac{g}{L}\times0.305\space L = 312\space g$ (rounded to 3 significant figures).