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Answer:
[tex]\rho_s=1.07g/cc^3[/tex]
Explanation:
From the question we are told that:
Mass [tex]M=5.29g[/tex]
Volume [tex]V=5.00mL=>5.0cc^3[/tex]
Generally the equation for Specific Gravity momentum is mathematically given by
[tex]Specific\ Gravity\ g= Density\ of\ Salt\ Solution\ \rho_s / Density\ of\ Water\ \rho_w[/tex]
[tex]g=\frac{ \rho_s} { \rho_w}[/tex]
[tex]\rho_s=\frac{ g} { \rho_w}[/tex]
[tex]\rho_s=\frac{ 5.29} { 5}[/tex]
[tex]\rho_s=1.07g/cc^3[/tex]
The specific gravity of this solution is = 1.06.
The specific gravity of a solution is the mass of a unit volume of a substance to the mass of a unit volume of a given reference material.
Therefore specific gravity of the solution can be calculated as follows:
[tex]sg = \frac{density. of . substance}{density . of . water} [/tex]
But, density of substance (salt): mass/volume
mass = 5.29g
volume = 5.00mL
Therefore density = 5.29/5
= 1.06g/ml
But density of water is = 1 g/ ml
Therefore, specific gravity = 1.06/1
= 1.06
The specific gravity of a solution is 1.06.
Learn more about specific gravity here:
https://brainly.com/question/543765