Use the General Gas Law
PV = nRT => n = PV/ RT
P= <span>10130.0 kPa
V= </span><span>50 L
R= </span><span>R = 8.314 L∙kPa/K∙mol
T= </span><span>300°C + 273 = 573 K
n = </span>10130.0 kPa 50 L / 8.314 L∙kPa/K∙mol <span>573 K
n = </span><span>106.32 mol</span>
Answer:- The pressure of ethanol would be 109 mmHg.
Solution:- This problem is based on Clausius clapeyron equation--
Given, = 63.5 + 273 = 336.5 K
= 34.9 + 273 = 307.9 K
= 400 mmHg
= ?
= 39.3 kJ/mol = 39300 J/mol
R = 8.314 J/mol.K
Let's plug in the values in the equation and do the calculations.
= 1.30
On taking anti ln to both sides...
=
= 3.67
= 400/3.67
= 109 mmHg
Volume of solution : = 3083.3 ml = 3.0833 L
<h3>Further explanation</h3>
Given
a 6.0% w/v
185 g of glucose
Required
Volume of solution
Solution
% w/v : the mass of solute in a volume of solution or 1 g solute in 100 ml of solution
Can be formulated :
% w/v = (mass solute(g) : volume solution(ml)) x 100
Mass of solute = mass of glucose = 185 g
The volume of solution :
= 185 g : 0.06
= 3083.3 ml
= 3.0833 L