question 28 (4 points)\nthe density of water at 30.0 °c is 0.995646 g/ml. a hydrometer gave the specific…

question 28 (4 points)\nthe density of water at 30.0 °c is 0.995646 g/ml. a hydrometer gave the specific gravity of a sugar solution at this temperature as 1.4567.\nthe density of the sugar solution is numerically __1__ and the units are __2__.\na. 4.3511×10^1 b. 4.8348×10^(-2) c. 4.3890×10^1 d. 4.8767×10^(-2)\ne. 1.4631 f. 6.8350×10^(-1) g. 1.4504 h. no units i. g\nj. g/ml k. ml/g l. ml

question 28 (4 points)\nthe density of water at 30.0 °c is 0.995646 g/ml. a hydrometer gave the specific gravity of a sugar solution at this temperature as 1.4567.\nthe density of the sugar solution is numerically __1__ and the units are __2__.\na. 4.3511×10^1 b. 4.8348×10^(-2) c. 4.3890×10^1 d. 4.8767×10^(-2)\ne. 1.4631 f. 6.8350×10^(-1) g. 1.4504 h. no units i. g\nj. g/ml k. ml/g l. ml

Answer

Explanation:

Step1: Recall density - specific gravity relationship

The formula for specific gravity (SG) is $SG=\frac{\rho_{substance}}{\rho_{reference}}$, where $\rho_{substance}$ is the density of the substance of interest and $\rho_{reference}$ is the density of the reference substance. Here, the reference substance is water at $30.0^{\circ}C$ with $\rho_{water}=0.995646\ g/mL$ and $SG = 1.4567$ for the sugar - solution.

Step2: Solve for the density of the sugar - solution

We can re - arrange the specific gravity formula to $\rho_{substance}=SG\times\rho_{reference}$. Substitute the given values: $\rho_{sugar - solution}=1.4567\times0.995646\ g/mL$. $\rho_{sugar - solution}=1.4567\times0.995646 = 1.4504\ g/mL$

Answer:

G. 1.4504, J. g/mL