Find the bioling point of a …
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Sia ? 6 years, 5 months ago
Given, wB = 0.520 g, wA = 80.2 g,
{tex}K_b = 0.52 \;{Km}^{-1}{/tex}
{tex}{M_2}({C_6}{H_{12}}{O_6}) = (6 \times 12 )+( 12 \times 1) +( 6 \times 16){/tex}
{tex}= \,180\,g\,mo{l^{ - 1}}{/tex}
{tex}\Delta {T_b} = \frac{{{K_b} \times {w_B} \times 1000}}{{{M_B} \times {w_A}}}{/tex}
{tex} = \frac{{0.52K{m^{ - 1}} \times 0.520g \times 1000g\,kg{\,^{ - 1}}}}{{180\,g\,mo{l^{ - 1}} \times 80.2g}}{/tex}
{tex}\Delta {T_b} = 0.0187K{/tex}
Boiling point of solution {tex}({{\rm{T}}_b}) = T_b^o + \Delta {T_b}{/tex}
{tex} = 373K + 0.0187K{/tex}
{tex}\Delta {T_b} = 373.019K{/tex}
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