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The boiling point of water at 735 Torr is 99.07 °C. The mass of $NaCl$ added in 100 g water $(K_{b} = 0.51 Kkg {mol}^{- 1})$ to make its boiling point 100 °C is

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10.68 g

5.34 g

2.67 g

26.7 g

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detailed solution

Correct option is B

$m = \frac{Δ T_{b}}{K_{b}} = \frac{(100 - 99.07)}{0.51} mol {kg}^{- 1}$

Assuming $NaCl$ to be completely dissociated, we will have

Molality of $NaCl = \frac{1}{2} (\frac{100 - 99.07}{0.51} mol {kg}^{- 1})$

Mass of $NaCl$ in 100 g water $= (\frac{1}{10}) \frac{1}{2} (\frac{0.93}{0.51}) (58.5) g = 5.33 g$

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The boiling point of water at 735 Torr is 99.07 °C. The mass of NaCl added in 100 g water Kb=0.51Kkgmol−1 to make its boiling point 100 °C is