81.4 g sample of ethyl alcohol contains 0.00 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The mass of pure ethyl alcohol is calculated by subtracting the mass of water from the total mass of the sample.Mass of pure ethyl alcohol $= 81.4

Vedclass · Accepted Answer

$81.4$

Vedclass · Answer

(A) The mass of pure ethyl alcohol is calculated by subtracting the mass of water from the total mass of the sample.Mass of pure ethyl alcohol $= 81.4 \ g - 0.002 \ g = 81.398 \ g$.According to the rules of significant figures for addition and subtraction,the result should be reported to the same number of decimal places as the measurement with the fewest decimal places.Here,$81.4 \ g$ has one decimal place,while $0.002 \ g$ has three decimal places.Therefore,the result must be rounded to one decimal place.Rounding $81.398 \ g$ to one decimal place gives $81.4 \ g$.

$81.4 \ g$ sample of ethyl alcohol contains $0.002 \ g$ of water. The amount of pure ethyl alcohol in $g$ to the proper number of significant figures is:

Similar Questions

$25^{\circ} C$ temperature in Fahrenheit and Kelvin scales are respectively:

The prefix $10^{18}$ is

Identify the carrier gas commonly used in gas chromatography.

Which of the following has the highest energy value?

One fermi is $....$ (in $cm$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module