An effusion container is filled with 2 l of an unknown gas. it takes 350 s for the gas to effuse into a vacuum. from the same container under the same conditions--same temperature and initial pressure, it takes 229 s for 2.00 l of o2 gas to effuse. calculate the molar mass (in grams/mol) of the unknown gas.

6.85g/mol

Explanation:

According to the Graham's law of diffusion,the rate of diffusion is inversely proportional to  the square root of molecular mass.

k=r√m.m

k=constant

r=rate of diffusion or effusion

m.m=molecular mass

taking ''1'' and '2" to be the two gases 1=oxygen,2=unknown gas

r1√m1=r2√m2,molar mass of oxygen=16

mathematically√m2=229√16/350

√m2=2.617

m2=6.85g/mol

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol