A certain mass of hydrogen gas collected over water at 10°C and 750mmHg pressure has a volume of 57cm3. Calculate the volume when it is dry at S. T.P ( S. V.P of water at 10°C = 9.2 mmHg .

The volume of gas when it is dry at S.T.P : 53.6 cm³

Given

P tot=750 mmHg

V₁=57 cm³

T₁=10 °C = 283 K

P₂ = 760 mmHg(STP)

T₂ = 273 K(STP)

Required

P dry gas at STP

Solution

Dalton's Law

P tot = P H₂O + P gas

P gas = 750 mmHg - 9.2 mmHg

P gas = 740.8 mmHg = P₁

Combined gas Law

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