Question

A 3.00-g sample of solid SnCl2·2H2O was heated such that the water turned to steam and was driven off. Assuming ideal behavior, what volume would that steam occupy at 1.00 atm and 100.0 °C?

Answer 1

Molar mass of SnCl2.2H2O = 1*MM(Sn) + 2*MM(Cl) + 4*MM(H) + 2*MM(O)

= 1*118.7 + 2*35.45 + 4*1.008 + 2*16.0

= 225.632 g/mol

mass of SnCl2.2H2O = 3.00 g

we have below equation to be used:

number of mol of SnCl2.2H2O,

n = mass of SnCl2.2H2O/molar mass of SnCl2.2H2O

=(3.0 g)/(225.632 g/mol)

= 1.33*10^-2 mol

moles of H2O = 2*moles of SnCl2.2H2O

= 2*1.33*10^-2 mol

= 2.66*10^-2 mol

we have:

P = 1.0 atm

n = 0.0266 mol

T = 100.0 oC

= (100.0+273) K

= 373 K

we have below equation to be used:

P * V = n*R*T

1 atm * V = 0.0266 mol* 0.08206 atm.L/mol.K * 373 K

V = 0.814 L

Answer: 0.814 L