Question

If the $6M\;KOH\;$ solution is $28\%$ by weight, then what is the specific gravity of this solution?
(A) $1.2$
(B) $2.4$
(C) $0.6$
(D) $1.8$

Answer 1

Specific gravity alludes to the proportion of the density of an item and the reference material. Besides, the particular gravity can advise us if the item will sink or buoy in reference material. Furthermore, the reference material is water that consistently has a density of 1 gram for each cubic centimeter or 1 gram for every millimeter.

Complete Step By Step Solution:
$28\% {{ }}\left( {w/w} \right)$ implies $28{{ }}gm{{ }}KOH$ in $100{{ }}gm$ arrangement.
In the $1L$ arrangement, there are $6$ moles of $KOH$ , for example $336{{ }}g$ .
Since, $KOH$ is $28\%$ by weight (as given), at that point
weight of water = $336 \times 72/28 = 18 \times 48 = 864g.$
Molarity= $w/Mwt{{ }} \times {{ }}1000/V\left( {ml} \right)$
$6 = {{ }}28/56{{ }} \times {{ }}1000/V\left( {ml} \right)$
$V = 1000/12 = 83.33ml$
Density= $100/83.33{{ }} = 1.2g/ml$
Explicit gravity= $1.2g/ml$
So, the answer is option A.

Additional Information:
Relative density, or explicit gravity, is the proportion of the density (mass of a unit volume) of a substance to the density of a given reference material. Explicit gravity for fluids is almost consistently estimated regarding water at its densest; for gases, the reference is air at room temperature. The expression "relative density" is regularly favored in logical use. In the event that a substance's overall density is short of what one, at that point it is less thick than the reference; in the event that more prominent than 1, at that point it is denser than the reference. In the event that the overall density is actually 1, at that point the densities are equivalent; that is, equivalent volumes of the two substances have a similar mass. In the event that the reference material is water, at that point a substance with a relative density (or explicit gravity) under 1 will coast in water.

Note:
Temperature and weight should be indicated for both the example and the reference. Weight is almost consistently $1{{ }}atm$ . Where it isn't, it is more common to determine the density straightforwardly. Temperatures for both example and reference differ from industry to industry. In English blending practice, the particular gravity, as determined above, is increased by $1000$ .