Answer:
[tex]\rho_s=\frac{\rho_w}{\frac{wt1}{SpGr1}+\frac{wt2}{SpGr2}}[/tex]
The only other data you need is the density of water ρw.
Explanation:
We can start by the volume balance
[tex]V_s=V_1+V_2[/tex]
We can replace the volumes with V=M/ρ
[tex]\frac{M_s}{\rho_s}=\frac{M_1}{\rho_1}+\frac{M_2}{\rho_2}[/tex]
If we divide every term by Ms
[tex]\frac{M_s/M_s}{\rho_s}=\frac{M_1/M_s}{\rho_1}+\frac{M_2/M_s}{\rho_2}[/tex]
By definition, wt=Mi/Msol, so we can replace that in the expression
[tex]\frac{1}{\rho_s}=\frac{wt1}{\rho_1}+\frac{wt2}{\rho_2}[/tex]
Then we have the expression of the density of the solution
[tex]\rho_s=\frac{1}{\frac{wt1}{\rho_1}+\frac{wt2}{\rho_2}}[/tex]
To replace ρ1 and ρ2, you have to multiply the specific gravity of the components and the density of water.
[tex]\rho_s=\frac{1}{\frac{wt1}{SpGr1\rho_w}+\frac{wt2}{SpGr2\rho_w}}\\\\\rho_s=\frac{\rho_w}{\frac{wt1}{SpGr1}+\frac{wt2}{SpGr2}}[/tex]
