ΔTb=Kb× molality

ΔTb=1K

Molality =\(={moles(n)\over weightofsolvent(Kg)}\)

Molality = \(= {1 X 1000 \over 50X50} = {10 \over 25} = {2\over5} \)

ΔTb=Kb× molality

1=Kb×\({2\over 5}\)

Kb=2.5 K Kg mol−1

\(\begin{equation} x = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cfrac{1}{a_4} } } } \end{equation}\)

\(A_{m,n} = \begin{pmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m,1} & a_{m,2} & \cdots & a_{m,n} \end{pmatrix}\)

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