# A CERTAIN HYDRATE HAS THE FORMULA MGSO4.XH2O.A QUANTITY OF 54.2g OF COMPOUND IS HEATED IN AN OVEN TO DRIVE OF THE WATER.IF THE STEAM GENERATED EXERTS THE PRESSURE OF 24.8ATM INA 2.0 L CONTAINER AT 120 C CALCULATE X ?

36 Points
13 years ago

Dear student,

P=24.8 bar

V = 2l = 0.002 m^3

T = 120 = 393K

PV = mRT/M => m = PVM/RT

=> m = 24.8*10^5*2*10^-3*18/8.314*10^3*393 =0.02732 kg = 27.32g

percent weight of water in the compound = (27.32/54.2)*100 = 50.4

molecular weight of compound = 24+32+64+x*18

=> 120+18x = 0.504*18x => x = 13.33 since x is natural number x is approximated to 13

hope this explains.

Gourav Chakraborty
13 Points
5 years ago
Given wt =54.2x
Molecular wt =120+18x
Pressure =24.8
Volume=2L
Temp.=393k
Applying ideal gas equation;
I.e pv=nRT
24.8 * 2 = given mass/ molar mass * 0.0821*393
24.8*2=54.2x / (120+18x) * 0.0821*393
After solving we get ;
X = 6.95 which is approx 7
Rudra Sohani
15 Points
4 years ago
Given mass of compound =54.2
Moles of compound=54.2/120+18x
One mole of compound will produce x mole of H2O
P=24.8
T=393K
V=2L
BY PV=nRT
Calculate n and equate it with 54.2x/(120+18x)
Find x