Answer : The mass of NaCl needed would be, 86.1 grams.
Explanation :
Formula used for Elevation in boiling point :
[tex]\Delta T_b=i\times k_b\times m[/tex]
or,
[tex]\Delta T_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]
where,
[tex]\Delta T_b[/tex] = change in temperature = [tex]1.500^oC[/tex]
[tex]k_b[/tex] = boiling point constant = [tex]0.5100^oC/m[/tex]
m = molality
i = Van't Hoff factor = 2 (for NaCl electrolyte)
[tex]w_2[/tex] = mass of solute (NaCl) = ?
[tex]w_1[/tex] = mass of solvent (water) = 1001 g
[tex]M_2[/tex] = molar mass of solute (NaCl) = 58.5 g/mol
Now put all the given values in the above formula, we get:
[tex]1.500^oC=2\times (0.5100^oC/m)\times \frac{w_2\times 1000}{58.5g/mol\times (1001g)}[/tex]
[tex]w_2=86.1g[/tex]
Therefore, the mass of NaCl needed would be, 86.1 grams.
The gram of NaCl needed to be added to increase the boiling temperature will be
Also know as boiling point, The boiling point of a liquid varies according to the applied pressure; the normal boiling point is the temperature at which the vapour pressure is equal to the standard sea-level atmospheric pressure.
Therefore,
[tex]\Delta T_b = ik_b(\frac{mass of NaCl}{molar mass of NaCl} * \frac{1000}{H_2O mass})[/tex]
[tex]\Delta T_b[/tex] = 1.5
i = 2
[tex]k_b[/tex] = 0.51
molar mass of NaCl = 58.5g/mol
Therefore.
[tex]1.5 = 2*0.51(\frac{mass of NaCl}{58.5} * \frac{1000}{1001})\\\\mass of NaCl = 86.115grams[/tex]
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