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```        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 ?```
8 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.
```
8 years ago
```							Given wt =54.2xMolecular wt =120+18xPressure =24.8Volume=2LTemp.=393kApplying ideal gas equation;I.e pv=nRT24.8 * 2 = given mass/ molar mass * 0.0821*39324.8*2=54.2x / (120+18x) * 0.0821*393After solving we get ; X = 6.95 which is approx 7
```
one year ago
```							Given mass of compound =54.2Moles of compound=54.2/120+18xOne mole of compound will produce x mole of H2OP=24.8T=393KV=2L BY PV=nRTCalculate n and equate it with 54.2x/(120+18x) Find x
```
5 months ago
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