Question

what weight of glucose would have to be be added to 1700g of water at 20°C to lower its vapour pressure by 0.001mm of hg. The vapour pressure of pure water is 17mm hg at 20°C.​

Answer 1

Answer: 1 gram

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=\frac{w_2M_1}{w_1M_2}$

where,

$p^o-p_s$ = lowering of vapor pressure = 0.001 mmHg

$p6)$ = vapor pressure of solvent  = 17 mmHg

$w_2$ = mass of solute  (glucose) = x g

$w_1$ = mass of solvent  = 1700 g

$M_1$ = molar mass of solvent (water) = 18 g/mole

$M_2$ = molar mass of solute (glucose) = 180 g/mole

Now put all the given values in this formula ,we get the vapor pressure of the solution.

$\frac{0.001}{17}=\frac{x\times 18}{1700\times 180}$

$x=1g$

Therefore, the weight of glucose that would have to be be added to 1700g of water is 1 gram.