Question

A solution of glucose is prepared with 0.052 g at glucose in 80.2 g of water. (Kf = 1.86 K kg mol-I and K kg mol-I )Mole fraction of glucose in the given solution is glucose then​

Answer 1

Answer:

80.148

Explanation:

hope it helps you

Answer 2

Given:

Mass of glucose $=0.052g$

Mass of water $=80.2g$

$\[{K_f} = 1.86\,Kkgmo{l^{ - 1}}\,and\,Kkgmo{l^{ - 1}}\,\]$

To Find: Mole fraction of glucose in the given solution of glucose

Solution:

Mole fraction of component A of a solution can be given as:

$\[{X_A} = \frac{{{n_A}}}{{{n_A} + {n_B}}}\]$

Where, $\[{{n_A}}\]$ and $\[{{n_B}}\]$ are the moles of componenets A and b respectively.

Number of moles of glucose can be given as:

$\[{{\text{n}}_{\text{A}}}{\text{ = }}\frac{{{\text{mass}}\,{\text{of}}\,{\text{glucose}}}}{{{\text{Molar}}\,{\text{mass}}\,{\text{of}}\,{\text{glucose}}}}\]$

$\[{{\text{n}}_{\text{A}}}{\text{ = }}\frac{{0.052g}}{{180\,g/mol}}\]$

$\[{n_A} = 0.00028\,mol\]$

Number of moles of water can be given as:

$\[{{\text{n}}_{\text{B}}}{\text{ = }}\frac{{{\text{mass}}\,{\text{of}}\,{\text{water}}}}{{{\text{Molar}}\,{\text{mass}}\,{\text{of}}\,{\text{water}}}}\]$

$\[{{\text{n}}_{\text{A}}}{\text{ = }}\frac{{80.2g}}{{18\,g/mol}}\]$

$\[{n_A} = 4.45\,mol\]$

Therefore, the mole fraction of glucose in the given solution can be given as:

$\[{X_A} = \frac{{0.00028}}{{0.00028 + 4.45}}\]$

$\[{X_A} = 6.29 \times {10^{ - 5}} \approx 6.3 \times {10^{ - 5}}\]$

Hence, the mole fraction of glucose in the given solution is $\[6.3 \times {10^{ - 5}}\]$.