Answer :
Answer : The molality of NaCl is, 1.95 mol/kg
Explanation :
Formula used for Elevation in boiling point :
[tex]\Delta T_b=i\times k_b\times m[/tex]
where,
[tex]\Delta T_b[/tex] = change in boiling point = [tex]2^oC[/tex]
[tex]k_b[/tex] = boiling point constant for water = [tex]0.512^oC/m[/tex]
m = molality
i = Van't Hoff factor = 2 (for electrolyte)
The dissociation [tex]NaCl[/tex] will be,
[tex]NaCl\rightarrow Na^++Cl^{-}[/tex]
So, Van't Hoff factor = Number of solute particles = [tex]Na^++Cl^{-}[/tex] = 1 + 1 = 2
Now put all the given values in the above formula, we get:
[tex]2^oC=2\times (0.512^oC/m)\times m[/tex]
[tex]m=1.95mol/kg[/tex]
Therefore, the molality of NaCl is, 1.95 mol/kg
The molality of NaCl is, 1.95 mol/kg
- The calculation is as follows;
[tex]\Delta T _ b = i \times k_b \times m\\\\2^{\circ}C = 2\times (0.512^{\circ} C/m) \times m[/tex]
m = 1.95 mol/kg
Here m represent the molarity
[tex]\Delta T_b[/tex] represents a change in boiling point i.e. [tex]2^{\circ}C[/tex]
[tex]k_b[/tex] represents the constant boiling point for water
i = Van't Hoff factor = 2 (for electrolyte)
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