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ICSE Class 10 Chemistry Question 17 of 72

Mole Concept & Stoichiometry Miscellaneous Exercises — Question 17

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Question 17

(a) Calculate the mass of substance 'A' which in gaseous form occupies 10 litres at 27°C and 700 mm pressure. The molecular mass of 'A' is 60.

(b) A gas occupied 360 cm3 at 87°C and 380 mm Hg pressure. If the mass of gas is 0.546 g, find it's relative molecular mass.

Answer

(a) Given, molecular mass of 'A' is 60

V1 = 10 L

T1 = 27 + 273 K = 300 K

P1 = 700 mm

T2 = 273 K

P2 = 760 mm

V2 = ?

Using formula:

P1V1T1\dfrac{\text{P}_1\text{V}_1}{\text{T}_1} = P2V2T2\dfrac{\text{P}_2\text{V}_2}{\text{T}_2}

Substituting in the formula,

700×10300\dfrac{700 \times 10}{300} = 760×V2273\dfrac{760 \times\text{V}_2}{273}

V2 = 700×10×273300×760\dfrac{700 \times10 \times 273}{300 \times760} = 1911228\dfrac{1911}{228} = 8.38 L

As, 22.4 L of A weighs 60 g

∴ 8.38 L of A will weigh 6022.4\dfrac{60}{22.4} x 8.38

= 22.446 = 22.45 g

(b) V1 = 360 cm3

T1 = 87 + 273 K = 360 K

P1 = 380 mm Hg pressure

T2 = 273 K

P2 = 760 mm Hg pressure

V2 = ?

Using formula:

P1V1T1\dfrac{\text{P}_1\text{V}_1}{\text{T}_1} = P2V2T2\dfrac{\text{P}_2\text{V}_2}{\text{T}_2}

Substituting in the formula,

380×360360\dfrac{380\times 360}{360} = 760×V2273\dfrac{760 \times\text{V}_2}{273}

V2 = 380×273760\dfrac{380 \times 273}{760} = 10,374760\dfrac{10,374}{760} = 136.5 cm3

136.5 cm3 of gas weigh = 0.546 g

22400 cm3 of gas weigh 0.546136.5\dfrac{0.546}{136.5} x 22400

= 89.6 amu