Question

‌How‌ ‌many‌ ‌grams‌ ‌of‌ ‌glucose‌ ‌(molecular‌ ‌mass=‌ ‌$180.9$g/mol)‌ ‌must‌ ‌be‌ ‌dissolved‌ ‌in‌ ‌255g‌ ‌
of‌ ‌water‌ ‌to‌ ‌increase‌ ‌the‌ ‌boiling‌ ‌point‌ ‌to‌ ‌${{102.36}^{\circ‌ ‌}}C$‌ ‌

Answer 1

Hint:The question is based on the concept of elevation of the boiling point. We know that the elevation in the boiling point is given by $\Delta {{T}_{b}}=\,i\times {{K}_{b}}\,\times \,m$. In this expression in place of molality we will simply write its formula and will deduce a formula for finding the mass of solute (here grams of glucose)

Complete step-by-step solution:Whenever we dissolve a solute in the solution its boiling point increases. And we know that the increase in boiling point of a solvent can be related to molality as;
$\Delta {{T}_{b}}=\,{{K}_{b}}\times m$ …$\left( i \right)$
 the term $i$=1 as glucose is non electrolyte
Here $m=$ molality of the solution and $\Delta {{T}_{b}}$= elevation in boiling point of solvent
We know molality $\left( m \right)$= $\dfrac{moles\,of solute}{mass\,of\,solvent\,\left( in\,kg \right)}$ …$\left( ii \right)$
Putting $\left( ii \right)\,in\,\left( i \right)$we get,
$\begin{align}
  & \Delta {{T}_{b}}=\,{{K}_{b}}\,\times \,\dfrac{moles\,of\,solute}{mass\,of\,solvent} \\
 & \Rightarrow \,moles\,of\,solute\,=\,\,\dfrac{\Delta {{T}_{b}}\,\times \,mass\,of\,solvent}{{{K}_{b}}} \\
 & \\
\end{align}$ …. $\left( iii \right)$
In this question solute is glucose, and solvent is water
$\Delta {{T}_{b}}\,=\,102.36\,-\,100\,=\,2.36{{\,}^{\circ }}C$
Mass of solvent (water) = $255g\,=\,\dfrac{255}{1000}=\,0.255kg$
${{K}_{b}}\,=\,0.52{{\,}^{\circ }}C\,kg/mol$ for water
Putting all these values in equation $\left( iii \right)$ we get,
moles of glucose $\begin{align}
  & \Rightarrow \,\dfrac{2.36{{\,}^{\circ }}C\,\times \,0.225\,kg}{0.52{{\,}^{\circ }}C\,kg/mol} \\
 & \, \\
\end{align}$
$\Rightarrow \,1.03\,moles$ of glucose.
We know that
Mass of solute= moles of solute $\times $ molar mass
Hence mass of glucose = moles of glucose $\times \,$ molar mass
Therefore, Mass of glucose$=\,1.03\,\times \,180.9=\,187.518$g or $188g$

Hence $188g$ of glucose must be dissolved in $255g$ of water to increase the boiling point to $102.36{{\,}^{\circ }}C$.

Note:The mass of the solvent in this type of question should be in kg. if it is in g, express it into kg. also note that the value of $i$ in the case of glucose will be one, as it is a nonelectrolyte. Always use an approximation method.