a certain hydrate has a formula MgSO4.xH2O.A quantity of 54.2g of the compound is heated in an oven to drive off the water. If the steam generated exerts a pressure of 24.8atm in a 2.0L container at 120°C, calculate x.
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Answered by
11
P = 24.8 bar
V = 2 l = 0.002 m ^ 3
T = 120 = 393 K
P V = m R T / M = > m = P V M / R T
= > m = 24.8 * 10 ^ 5 * 2 * 10 ^ - 3 * 18 / 8.314 * 10 ^ 3 * 393 = 0.02732 kg = 27.32 g
Percent mass of water in the compound = ( 27.32 /54.2 ) * 100 = 50.4
Molecular mass of compound = 24 + 32 + 64 + x * 18
= > 120 + 18 x = 0.504 * 18 x = > x = 13.33 since x is natural figure x is approximated to 13
Answered by
19
Answer:7
Explanation:no of moles of h2o=PV/RT
=24.8×2.0/0.0821×393=1.54mol
mass of h2o produced=1.54mol×18g h2o/1molh2o=27.69g
mass of hydrate is sum of mass of mgso4 and h2o
mass of mgso4 produced=26.5g
no of moles of mgso4=26.5g mgso4×1mol mgso4/120g mgso4
=0.2202
x=no of moles of h2o/no of moles of mgso4=1.537/0.2202 =7
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