Question

1.82 g. of glucose (molar mass-180) is dissolved in 25g of water. Calculate (a) the molality (b) mole fraction of glucose and water.

Answer 1

Given :

• mass of glucose = 1.82 g
• molar mass of glucose = 180 g
• mass of water = 25 g
• molar mass of water = 18 g

To find :

• molality
• mole fraction of glucose
• mole fraction of water

Solution :

$\displaystyle\bigstar \: \underline \mathtt \blue{molality}$

we know that ,

$\boxed{ \mathtt{m = \frac{moles \: of \: solute}{mass \: of \: solvent(kg)} }}$

Solute = glucose

Solvent = water

First we need to find moles of glucose ,

we know that ,

$\displaystyle\implies \mathtt{n_g = \frac{mass_g}{molar \: mass_g} }$

$\displaystyle\implies\mathtt{n_g = \frac{1.82}{180} }$

$\displaystyle\implies {\mathtt{n _g= 0.01}}$

Substituting the above value in the equation ,

$\displaystyle\implies \mathtt{m = \frac{0.01 \times 1000}{25} }$

$\displaystyle\boxed{ \red{\mathtt{m = 0.4 \: molal}}}$

$\bigstar \underline\mathtt \blue {mole \: fraction \: of \: glucose}$

we know that ,

$\boxed{\mathtt{x_g = \frac{n_g}{n_g + n_w} }}$

$\displaystyle\implies \mathtt{n _g= 0.01}$

$\displaystyle\implies \mathtt{n _w= \frac{25}{18} }$

$\displaystyle\implies \mathtt{n _w= 1.38}$

Now substituting the values of n g and n w in the above equation ,

$\displaystyle\implies \mathtt{x_g = \frac{0.01}{0.01 + 1.38} }$

$\displaystyle\implies \mathtt{x_g = \frac{0.01}{1.39}}$

$\displaystyle\boxed{ \red{\mathtt{x_g = 0.01}}}$

$\bigstar \underline\mathtt \blue {mole \: fraction \: of \: water}$

we know that ,

$\boxed{\mathtt{x_w= \frac{n_w}{n_g + n_w} }}$

Now substituting the values of n g and n w in the above equation ,

$\displaystyle\implies \mathtt{x_w = \frac{1.38}{0.01 + 1.38} }$

$\displaystyle\implies \mathtt{x_w = \frac{1.38}{1.39}}$

$\displaystyle\boxed{ \red{\mathtt{x_w = 0.9}}}$