Depression is freezing point is a colligative property. It is mathematically expressed as ΔTf = Kf X m
where Kf = <span>freezing point depression constant = 1.86°c kg /mol (for water)
m = molality of solution = 1.40 m
</span>∴ ΔTf = Kf X m = 1.86 X 1.40 = 2.604 oC
Now, for water freezing point = 0 oC
∴Freezing point of solution = -2.604 oC
m = given mass of gas = 3.82 g
M = molar mass of gas = ?
T = temperature of laboratory = 302 K
P = air pressure = 1.04 atm = 1.04 x 101325 pa
V = volume of gas = 0.854 L = 0.854 x 10⁻³ m³
using the ideal gas equation
PV = (m/M) RT
inserting the above values
(1.04 x 101325) (0.854 x 10⁻³) = (3.82/M) (8.314) (302)
M = 106.6 g
hence the molar mass of the gas comes out to be 106.6 g
Answer:
Explanation:
Atomic mass of element will be weighted average atomic mass of the element
= 89.90470 x 0 .5293 + 90.90565 x. 1154 + 91.90504 x 0. 1765 + 93.90632 x 0.1788
= 47.58955 + 10.490 + 16.2212 + 16.790
= 91.10 u .
