Class 12th

${\lambda }_m^{\infty }$  of Nal = 12.7 mSm² mol^-1

${\lambda }_m^{\infty }$ of NaNO₃ = 12.0 m Sm² mol^-1

${\lambda }_m^{\infty }$ of AgNO₃ = 13.3 mSm² mol^-1

$\therefore {\lambda }_{Agl}^{\infty }=\left[{\lambda }_m^{\infty }\left(AgN{O}_3\right)+{\lambda }_m^{\infty }\left(Nal\right)\right]-{\lambda }_m^{\infty }NaN{O}_3$                

= (13.3 + 12.7) - 12.0 = 14.0 m Sm² mol^-1

  $\Delta T_b=i*K_b*m$

 or $\Delta T_b=K_b*m$   (i = 1 for non-electrolyte)

  $or\frac{{\left(\Delta T_b\right)}_A}{{\left(\Delta T_b\right)}_B}=\frac{{\left(K_b\right)}_A}{{\left(K_b\right)}_B}$

  $or\frac{x}{y}=\frac{1}{8}\left|\therefore y=8\right|$              

In Fe_0.93O

Let us suppose that fraction of Fe²⁺ is x and fraction of Fe³⁺ is (1 – x)

$x*\left(+2\right)+\left(1-x\right)*\left(+3\right)=$ O.S. of Fe atom

$or2x+3-3x=\frac{+200}{93}$

$\therefore \mathrm{\%}ofF{e}^{2+}=85\mathrm{\%}and\mathrm{\%}ofF{e}^{3+}=15\mathrm{\%}$

In Fe_0.93O

Let us suppose that fraction of Fe²⁺ is x and fraction of Fe³⁺ is (1 – x)

$x*\left(+2\right)+\left(1?x\right)*\left(+3\right)=$ O.S. of Fe atom

$or\text{?}\text{?}2x+3?3x=\frac{+200}{93}$

$?\mathrm{\%}\text{?}\text{?}of\text{?}\text{?}F{e}^{2+}=85\mathrm{\%}\text{?}\text{?}and\text{?}\text{?}\mathrm{\%}\text{?}\text{?}of\text{?}\text{?}F{e}^{3+}=15\mathrm{\%}$

Consider the image below

Consider the image given below

Based upon the properties and uses of chemical substances.

The minimum marks requirement to get admission in BSc MRT in KUHS is 50% in 12th PCB with 95% marks in 12th, you seem to have a good chance of getting admitted to this course.