Question

9 g of glucose (mol wt=180) is dissolved in 90 g of H₂O. Relative lowering of vapour pressure is​

Answer 1

Answer:

Relative lowering of vapour pressure = Mole fraction of solute ie glucose.

Explanation:

Number of moles of glucose = 9/180 = 1/20.

Number of moles of water = 90/18 = 5.

Therefore mole fraction of glucose=

$\frac{ \frac{1}{20} }{ \frac{1}{20} + \frac{100}{20} }$

=

$\frac{1}{101}$

= 0.099.

* You can round it of to 0.01.