the reaction for equation is written as follows
2Cs (s) + 2H2O(l) = 2CsOH (aq) + H2 (g)
The mass of cesium is calculated as follows
Find the moles of H2 by use of ideal gas equation that is Pv = nRT
where n is the number of moles
n = PV/RT
P= 48.1 ml in liters = 48.1 /1000= 0.0481 l
T= 19 + 273.15 = 292.15K
P= 768mm hg
R (gas Constant)= 62.364 l.mmhg/k.mol
n= ( 768mmhg x0.0481 L) /( 62.364 L.mm hg/k.mol x 292.15k) = 2.028 x10^-3 moles
by use of mole ratio from reacting equation between Cs to H2 which is 2 :1 the moles of Cs is therefore = ( 2.028 x10^-3) x 2 = 4.05 x10^-3 moles
mass of cs is therefore = moles x molar mass of cs( 132.9g/mol)
=( 4.05 x10^-3)mol x 132.9 g/mol = 0.539 grams
