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`        The molal elevation constant of water is 0.51 Km–1 The boiling point of 0.1 molal aqueous NaCl solution is nearly :Options1.  100.05°C2.  100.1°C3.  100.2°C4.  101.0°C`
10 months ago

Chhavi Jain
92 Points
```							Dear student ,  Use the formula :                                       ΔTb = Kb m                                    Tb –  Tbo  = 0.51 * 0.1 =0.051                                     Since water boils at 373.15 K  , So  Tbo = 373.15 K                                 Therefore ,  Tb - 373.15 = 0.051                                                                             Tb  = 373.201  K (  bp of aq. NaCl sol.)Chemistry ConsultantAskiitians
```
10 months ago
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### Course Features

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• Test paper with Video Solution
• Mind Map
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• Discussion Forum
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