In a certain experiment, magnesium boride (Mg_3B_2) reacted with acid to form a

Question

In a certain experiment, magnesium boride (Mg_3B_2) reacted with acid to form a mixture of four boron hydrides (B_x H_y), three as liquids (labeled I, II, and III) and one as a gas (IV). When a 0.1000-g sample of each liquid was transferred to an evacuated 750.0-mL container and volatilized at 70.00 degree C, sample I had a pressure of 0.05951 atm; sample II, 0.07045 atm; and sample III, 0.05767 atm. What is the molar mass of each liquid? Boron is 85.63% by mass in sample I, 81.10% in II, and 82.98% in III. What is the molecular formula of each sample? Sample IV was found to be 78.14% boron. Its rate of effusion was compared to that of sulfur dioxide; under identical conditions, 350.0 mL of sample IV effused in 12.00 min and 250.0 mL of Sulfur dioxide effused in 13.04 min. What is the modular formula of sample IV?

Explanation / Answer

(a) As the liquid samples are volatilized, consider the ideal gas law

PV=nRT

where P - Pressure in atm

V- Volume in L

n - number of moles

R - gas constant, 0.082 057 46 L atm K1 mol1

T - Temperature in K

Given Mass = 0.1 g

Volume = 750 mL = 0.75 L

T = 70+273.15= 343.15 K

n=PV/RT, Molar mass =Mass/moles

For Sample I

P= 0.05951 atm

n = (0.0591*0.75)/(0.08205746*343.15)

= 0.00157 mol

Molar Mass = 0.1/0.00157 = 63.52 g/mol

For Sample II

P= 0.07045 atm

n = (0.07045*0.75)/(0.08205746*343.15)

= 0.00187 mol

Molar Mass = 0.1/0.00187 = 53.36 g/mol

For Sample III

P= 0.05767 atm

n = (0.05767*0.75)/(0.08205746*343.15)

= 0.00153 mol

Molar Mass = 0.1/0.00157 = 65.18 g/mol

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